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of  the 

University  of  California 

Los  Angeles 

Form  L   I 

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This  book  is  DUE  on  the  last  date  stamped  below 


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MATHEMATICS 

FOR 

AGRICULTURAL  STUDENTS 


McGraw-Hill  BookGompany 


Electrical  World         The  Engineering  andMining  Journal 
Engineering  Record  Engineering  News 

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Metallurgical  and  Chemical  Engineering  Power 


2j_ 


f 


EDITED  BY  CHARLES  S.  SLIGHTER 

MATHEMATICS 

FOR 

AGRICULTURAL  STUDENTS 

BY 
HENRY  C.  WOLFF,  PH.  D. 


ASSISTANT    PROFESSOR   OF  MATHEMATICS 
UNIVERSITY   OP    WISCONSIN 


FIRST  EDITION 


McGRAW-HILL  BOOK  COMPANY,  INC. 

239  WEST  39TH  STREET,  NEW  YORK 

6  BOUVERIE  STREET,  LONDON,  E.  C. 

1914 


COPYRIGHT,  1914,  BY  THE 
McGRAW-HiLL  BOOK  COMPANY,  INC. 


THE. MAPLE. PKES3. YORK-PA 


£3 


PREFACE 

The  present  book  was  designed  as  the  text  for  a  working  course 
in  elementary  Mathematics  given  in  the  College  of  Agriculture  of 
the  University  of  Wisconsin.  It  is  not  a  text  book  of  Advanced 
Algebra,  Trigonometry,  Analytic  Geometry,  or  "Practical  Mathe- 
matics." Students  pursuing  a  scientific  course,  in  which  but  a 
year  of  college  Mathematics  is  offered,  receive  little  profit  from  a 
formal  course  in  Advanced  Algebra,  Trigonometry,  or  Analytic 
Geometry.  On  the  other  hand,  such  students  benefit  little  from  a 
so-called  "practical"  course  of  a  type  which  does  little  more  than 
train  them  to  substitute  in  given  formulas,  or  to  use  formulas 
merely  as  a  means  to  an  end.  Such  a  course  does  little  to  develop 
the  habit  or  power  of  clear  and  logical  thinking. 

This  book  is  the  outgrowth  of  mimeograph  notes  used  for  the 
past  three  or  four  years  with  undergraduate  students  of  agricul- 
ture. With  the  exception  of  the  illustrations  and  exercises,  the 
book  contains,  nevertheless,  nothing  which  is  of  exclusive  in- 
terest to  agricultural  students.  It  may  be  used  equally  well 
with  any  class  of  scientific  students  who  desire  only  a  short 
course  in  mathematics  beyond  elementary  Algebra  and  Geometry. 

The  plan  of  the  book  is: 

First,  to  select  material  primarily  on  the  basis  of  its  usefulness 
to  scientific  students. 

Second,  to  illustrate  the  principles  with  problems  of  interest  to 
the  agricultural  student,  or  with  problems  having  direct  applica- 
tion to  his  work. 

Third,  to  give  minute  and  detailed  explanations  of  all  new 
work,  and  to  assume  the  minimum  attainments  in  mathematical 
preparation  on  the  part  of  the  student.  On  the  other  hand,  details 
are  often  omitted  from  work  that  is  not  new  to  the  student,  in 
order  that  they  may  be  supplied  by  the  student  himself. 

Fourth,  to  give  detailed  directions  for  doing  work  "at  home"  so 


vi  PREFACE 

that  the  student  of  moderate  mathematical  attainments  may, 
nevertheless,  "learn  by  doing." 

Fifth,  to  include  certain  material  not  elsewhere  readily  acces- 
sible to  the  student  of  science,  for  the  purpose  of  rendering  the 
book  useful  for  reference  throughout  the  four  years  of  the  college 
course. 

Sixth,  to  select  the  topics  and  the  amount  of  material  under 
each  topic  so  that  either  a  half  year  or  a  full  year  may  be 
devoted  to  the  work. 

In  writing  this  book  the  author  had  in  mind  the  preparation 
of  a  text  which  would : 

First,  train  the  student  to  do  neat  and  careful  work. 

Second,  encourage  the  student  to  make  further  use  of  elemen- 
tary Algebra  and  Geometry. 

Third,  develop  in  the  student  the  habit  of  careful  and  logical 
thinking. 

Fourth,  train  the  student  to  study  a  problem  with  a  view  of 
discovering  the  shortest  and  easiest  method  of  handling  it,  rather 
than  attacking  it  by  the  "first  thought-of "  method. 

Fifth,  show  the  student,  by  illustrations  and  exercises,  how 
mathematics  may  be  helpful  in  pursuing  other  subjects  of  study 
••and  useful  in  a  "practical"  way. 

While  this  book  includes  enough  material  for  a  year's  course  of 
study,  it  is  believed  that  it  does  not  contain  too  much  material 
for  a  shorter  course.  The  inclusion  of  additional  material  in  the 
text,  will  at  least  let  the  student  know  of  the  existence  of  sub- 
jects in  mathematics  not  covered  by  him  in  the  class  room. 
Even  this  superficial  knowledge  may  be  very  helpful  to  him 
later  in  connection  with  other  scientific  work,  especially  if  he, 
knows  he  can  find  brief  discussions  in  a  familiar  book. 

The  author  takes  this  opportunity  to  thank  Professor  C.  S. 
Slichter  for  assistance  in  the  preparation  of  the  entire  manuscript; 
and  to  acknowledge  his  indebtedness  to  Professor  E.  V.  Hunting- 
ton  for  valuable  suggestions  and  criticisms.  Acknowledgments 
are  due  Mr.  E.  Taylor  and  Mr.  T.  C.  Fry  for  suggestions  based 
upon  their  use  of  the  preliminary  notes  in  the  class  room. 

A  very  brief  review  of  elementary  Algebra  is  given  in  the  intro- 
duction, besides  a  list  of  materials  and  instruments.  In  the 


PREFACE  vii 

appendix  is  given,  for  reference,  a  list  of  common  mathematical 
symbols  and  a  few  formulas  of  mensuration. 

Material  in  the  book  which  may  be  omitted  when  less  than  a 
full  year  is  devoted  to  the  course,  is  indicated  by  enclosing 
exercise  and  section  numbers,  and  chapter  headings  within 
brackets. 

HENRY  C.  WOLFF. 
UNIVERSITY  OP  WISCONSIN 
September,  18,  1914 


CONTENTS 

PAGE 

PREFACE     v 

INTRODUCTION 1 

CHAPTER 

I     GRAPHIC  REPRESENTATION 23 

II    LOGARITHMS 60 

III  THE  CIRCULAR  FUNCTIONS:  THE  TRIANGLE 80 

IV  THE  ELLIPSE 128 

V     THE  SLIDE  RULE 143 

VI     STATICS 155 

VII     PERMUTATIONS,  COMBINATIONS,  AND  THE  BINOMIAL  EX- 
PANSION.   189 

VIII     PROGRESSION 200 

IX     PROBABILITY 207 

X     SMALL  ERRORS 231 

XI     POINT,  PLANE,  AND  LINE  IN  SPACE 239 

XII     MAXIMA  AND  MINIMA 252 

XIII     EMPIRICAL  EQUATIONS 264 

APPENDIX 286 

INDEX                                                                                                   .  303 


MATHEMATICS  FOR 
AGRICULTURAL  STUDENTS 


INTRODUCTION 

1.  Instruments  and  Materials.  A  large  number  of  problems  in 
this  book  are  to  be  solved  graphically  or  solved  both  graphically 
and  analytically,  and  a  large  number  of  the 
exercises  are  exercises  in  graphic  representation 
of  observed  or  computed  data.  In  order  that 
this  work  may  be  performed  neatly  and  accur- 
ately, it  is  necessary  that  the  student  have  at 
least  a  few  simple  drawing  instruments.  The 
indispensable  instruments,  together  with  other 
material  required  for  the  work,  are  here  listed 
and  briefly  described. 

1.  Two  4H  Drawing  Pencils.  One,  sharp- 
ened to  a  fine  point,  is  used  for  marking  points 
upon  paper,  or  for  sketching  free-hand.  The 
other,  sharpened  to  a  chisel-point,  is  used  for 
drawing  straight  lines.  Two  views  of  a  chisel- 
pointed  pencil  are  given  in  Fig.  1.  When 
drawing  lines,  a  flat  side  of  the  chisel-point  is 
held  against  the  straight  edge  of  the  guide, 
thus  causing  the  pencil  always  to  move  in  a 
direction  with  the  chisel  edge  and  to  produce  a 
fine  and  distinct  line.  For  fine  and  accurate 
work,  both  pencil  points  must  be  kept  sharp. 
Neither  point  ought  to  become  so  rounded  or 
flat  that  any  appreciable  metallic  luster  will 
appear  at  the  wearing  surface  of  the  graphite. 


FIG.  1.— Two 
views  of  a  chisel- 
pointed  pencil. 


A  good  way  of 

keeping  pencil  points  sharp  is  by  occasionally  whetting  upon  fine 

1 


2  MATHEMATICS  [§1 

sand-paper.  Small  pads  of  sand-paper  for  this  purpose  may  be 
procured  from  stationers  or  dealers  in  drawing  materials.  Hex- 
agonal pencils  should  always  be  used.  The  student  may  prefer 
to  have  one  pencil  sharpened  at  both  ends;  one  end  with  a  round 
point,  and  the  other  end  with  a  chisel-point. 

2.  One  3H  Pencil,  sharpened  to  a  round  point  for  lettering. 

3.  A  small  drawing  board  made  of  soft  wood  to  which  the  draw- 
ing paper  may  be  fastened  with  thumb-tacks.    A  drawing  board 
should  have  at  least  one  straight  edge  along  which  the  T-square 
may  slide.     A  board  12  by  18  inches  will  be  large  enough. 


FIG.  2. — T-square. 

4.  A  small  T-square  about  15  inches  long.    A  T-square,  Fig. 
2,  consists  simply  of  a  thin,  straight-edged  blade,  A,  screwed  to  a 
heavier  block  of  wood,  the  head,  B.    The  inside  of  the  head  has  a 
smooth,  straight  edge.    The  T-square  is  placed  upon  the  drawing 
board  with  its  head  held  against  the  left-hand  edge  of  the  drawing 
board.    By  sliding  it  up  and  down  with  the  left  hand,  the  draughts- 
man is  enabled  to  draw  with  it  as  guide  a  series  of  parallel  lines 
running  from  left  to  right.    From  this  use  of  the  T-square,  it  is 
seen  that  the  edge  of  the  drawing  board  along  which  it  slides 
must  be  smooth  and  straight,  otherwise  the  lines  drawn  along  the 
blade  of  the  square  would  not  be  parallel.    It  is  hot  at  all  necessary 
that  the  blade  of  a  T-square  be  exactly  at  right  angles  to  the  inner 
edge  of  the  head. 

5.  A  60°  and  a  45°  Transparent  Celluloid  Triangle.    Drawing 
triangles  are  thin  triangular  guides,  usually  made  of  wood  or 
celluloid,  one  angle  of  which  is  a  right  angle.    (See  Fig.  3.) 


§1]  INTRODUCTION  3 

In  a  60°  triangle,  the  acute  angles  are  60°  and  30°,  and  in  a 
45°  triangle,  the  two  acute  angles  are  each  45°. 

With  these  triangles  straight  lines  may  be  drawn  making  angles 
of  15°,  30°,  45°,  60°,  75°,  and  90°  with  the  lines  drawn  with  the  T- 
square  as  a  guide. 

For  the  work  in  this  book  6-inch  triangles  will  be  found  large 
enough,  although  an  additional  larger  60°  triangle  will  be  found 
very  convenient. 


60  Triangle  45  Triangle 

FIG.  3.— 60°  and  45°  triangles. 

6.  A  protractor  is  an  instrument  for  laying  off  or  measuring 
angles  upon  a  drawing.     One,  consisting  of  angular  graduations 
upon  a  semicircular  piece  of  transparent  celluloid,  about  6  inches 
in  diameter,  is  recommended  for  this  work.     (See  Fig.  4.) 

7.  A  scale  divided  decimally  (the  unit  of  length  divided  into 
tenths)  should  be  used  rather  than  one  in  which  the  unit  is  divided 
into  eighths,  twelfths,  or  sixteenths.   It  is  advised  that  a  triangular 
box-wood  scale  about  4  inches  long,  or  longer,  be  used  in  this  work. 
(See  Fig.  5.) 

8.  A  pair  of  6-inch  pencil  compasses  used  for  drawing  circles 
and  arcs  of  circles.     The  lead  in  a  cheap  pair  of  compasses  usually 
is  not  found  satisfactory  and  should  be  replaced  by  a  good  piece 
of  medium  hardness,  sharpened  to  a  narrow  chisel-point.    The 
lead  should  be  so  adjusted  that  when  the  two  legs  of  the  compasses 
are  brought  together  the  two  points  will  be  together  or  directly 
opposite  each  other. 


4  MATHEMATICS  [§1 

9.  A  good  rubber  eraser  for  erasing  lead  pencil  marks. 

10.  Twelve  small  thumb-tacks. 

11.  Fifty  sheets  of  squared  paper,  form  Ml,  used  for  plotting 
data. 


FIG.  4. — Protractor. 

12.  Twelve  sheets  of  paper,   form   M7,  used  for  numerical 
calculations  and  for  tables. 

13.  Six  sheets  of  polar  coordinate  paper,  form  M3. 

14.  Twenty-four  sheets  of  heavy  bond  paper,  letter  size,  8£  by 
11  inches;  to  be  used  for  drawings. 


Y 


\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\V 

FIG.  5. — Triangular  scale. 


\ 


15.  Two  heavy  paper  note  covers  suitable  for  holding  paper 
85  by  11  inches  without  folding.    These  covers  are  to  be  used  for 
submitting  daily  work  to  the  instructor. 

16.  One  note  cover  large  enough  for  filing  8J  by  11-inch  sheets 


§2]  INTRODUCTION  5 

without  folding.  This  cover  is  to  be  used  for  filing  the  written 
work  after  having  been  examined  and  returned  by  the  instructor. 

17.  A  set  of  "Slichter's  four-place  logarithmic  and  trigono- 
metric tables"  used  in  connection  with  numerical  computations. 

There  are  other  logarithmic  and  trigonometric  tables  published, 
but  in  order  to  do  efficient  classroom  work  without  waste  of  time, 
there  must  be  complete  uniformity  in  the  tables  used  by  the  mem- 
bers of  the  class. 

If  the  student  wishes  larger  tables  for  his  own  private  use,  the 
two  following  are  among  the  best: 

Albrecht's  five-place,  and 
Bremiker's  six-place. 

While  the  above  list  of  instruments  and  materials  includes  about 
the  minimum  in  number  and  smallest  in  size  for  efficient  work, 
more  and  better  instruments  may  be  purchased  if  the  student 
desires. 

2.  General  Directions.  All  work  must  be  done  neatly.  In 
general,  accuracy  in  graphic  work  can  only  be  acquired  by  neat 
work.  At  first,  perhaps,  much  time  will  be  sacrificed,  but  after 
the  habit  of  doing  neat  work  has  been  established  much  time  will 
be  saved  both  in  avoiding  errors  and  in  the  ease  with  which  errors 
may  be  located  or  the  work  checked.  The  student  at  the  begin- 
ning must  not  be  discouraged  if  a  large  number  of  his  drawings  and 
exercises  are  returned  marked  "  redraw"  or  "rewrite,"  for  neatness 
of  work  will  always  be  insisted  upon  as  one  of  the  means  o*  obtain- 
ing accurate  work. 

Unless  special  directions  are  given,  all  work  submitted  to  the 
instructor  is  to  be  upon  paper  of  letter  size,  8£  by  11  inches,  and 
enclosed  without  folding  in  a  paper  folder.  These  exercises  when 
returned  by  the  instructor  are  to  be  corrected  in  accordance  with 
instructions,  and  kept  in  order  in  a  note  cover.  Exercises 
marked  "rewrite"  or  "redraw  "are  to  be  done  over,  and  both 
the  original  and  the  corrected  sheets  filed.  This  file  of  exercises 
may  be  called  for  at  any  time  by  the  instructor. 

All  drawings  are  to  be  done  in  pencil,  unless  the  student  has  had 
training  in  the  use  of  the  ruling  pen,  in  which  case  he  may,  if  he 
desires,  "ink  in"  his  drawings  with  drawing  ink. 


6  MATHEMATICS  [§3 

All  descriptive  work  upon  a  drawing  is  to  be  lettered,  not 
written.1  Each  drawing  or  exercise  handed  in  must  be  numbered 
in  the  upper  right-hand  corner.  It  must  have  a  complete  yet 
brief  title,  and  contain  a  sufficient  number  of  descriptive  words 
in  the  body  of  the  drawing  so  that  anyone  familiar  with  the  prob- 
lem will  be  able  to  read  the  drawing;  i.e.,  to  understand  the 
meaning  of  every  line  or  figure  in  the  drawing. 

ABCDEFGHIJK 


a  b  cd  efghijklmno 
pqrstuv  wxyz 

1 234  567890 

FIG.  6. — Set  of  letters  and  figures. 

All  exercises — not  drawings — are  to  be  done  in  ink.  Use  any 
good  writing  fluid.  It  is  suggested  that  the  student  acquire 
the  habit  of  working  numerical  problems,  and  the  like,  with 
pen  and  ink. 

3.  To  draw  a  straight  line  through  a  point  with  the  T-square 
or  triangle  as  guide,  place  the  triangle  or  T-square  upon  the  paper, 
so  that  the  point  is  just  visible  at  one  of  its  edges;  press  the  tri- 
angle firmly  against  the  paper  with  the  fingers  of  the  left  hand  and 
with  the  pencil  in  the  right  hand  hold  a  flat  side  of  the  chisel-point 
against  the  straight  edge  and  move  the  pencil  from  left  to  right. 
The  pencil  should  be  held  with  its  upper  end  tipped  slightly  away 

1  A  set  of  letters  and  figures  is  given  in  Fig.  6. 


§3] 


INTRODUCTION 


from  the  triangle  in  order  to  bring  the  point  of  the  pencil  in  contact 
with  the  straight  edge.  (See  Fig.  7.)  As  far  as  possible,  the  whole 
right  hand  should  be  given  a  straight-line  motion  in  order  that  the 
pencil  may  be  made  to  move  parallel  to  itself. 


(Pencil 


Pencil 


Triangle 


Triangle 


FIG.  7. — a,  Incorrect  way  of  holding  a  pencil;  b,  correct  way  of  holding 

a  pencil. 

Care  must  be  taken  that  the  triangle  is  well  pressed  down  upon 
the  paper  at  the  point  where  the  line  is  being  drawn.  This  is 
partly  accomplished  by  the  pressure  of  the  fingers  of  the  left  hand, 
and  partly  by  pressing  down  upon  the  triangle  with  the  third  and 


fourth  fingers  of  the  right  hand.  This  pressing  down  with  the 
fingers  of  the  right  hand  not  only  helps  to  hold  the  triangle  in 
place,  but,  at  the  same  time,  gives  a  rest  to  the  hand. 


8  MATHEMATICS  [§4 

4.  Parallel  Lines.  Horizontal  parallel  lines  may  be  drawn  with 
the  aid  of  the  T-square. 

Parallel  lines  making  angles  of  30°,  45°,  60°  or  90°  with  the 
horizontal  may  be  drawn  with  the  aid  of  triangles  placed  upon  the 
edge  of  the  T-square. 

Parallels  other  than  those  mentioned  above  may  be  drawn  as 
follows:  Suppose  that  through  the  point  P,  Fig.  8,  we  wish  to 
draw  a  line  parallel  to  the  line  AB.  Place  two  triangles  together, 
as  shown  in  the  figure,  one  edge  of  No.  2  coinciding  with  the  given 
line.  Hold  triangle  No.  1  fixed  and  slide  No.  2  along  it  in  the 
direction  of  the  arrow  until  its  edge  which  originally  coincided  with 
AB  passes  through  the  point  P.  Along  this  edge  draw  a  line,  PQ. 
PQ  is  parallel  to  AB,  for,  by  geometry,  if  two  straight  lines  are  cut 
by  a  transversal,  and  if  the  interior  exterior  angles  on  the  same 
side  of  the  transversal  are  equal,  the  lines  are  parallel. 

Exercises 

1.  Fasten  a  sheet  of  paper  upon  the  drawing  board  with  its  longer 
dimension  extending  from  left  to  right.     About  2  inches  from  the 
lower  edge  of  the  sheet  draw,  with  the  aid  of  the  T-square,  a  horizontal 
line  6  inches  long.     Divide  this  line  into  six  equal  parts.     At  the 
points  of  division  draw  vertical  lines  with  the  aid  of  the  T-square 
and  triangle.     Beginning  with  the  horizontal  line  mark  off  four  equal 
spaces,  each  1  inch  long,  upon  the  left-hand  vertical  line.     From  the- 
points  of  division  just  located  draw  horizontal   lines,  thus   forming 
a  rectangle  4  by  6  inches,  divided  into  twenty-four  equal  squares. 
Through  the    lower  left-hand   corner  draw  lines  making  angles    of 
30°,  45°,  and  60°  with  the  horizontal.      Divide  the  upper  left-hand 
square  into  100  equal  squares.     A  drawing  (reduced)  of  this  exercise 
is  given  in  Fig.  9.     Complete  the  exercise  by  giving,  to  two  decimal 
places,  the  dimensions  and   quotients  omitted  in  Fig.  9. 

2.  In  a  45°  triangle,  prove  by  geometry  that  the  hypotenuse  is  \/2 
times  as  long  as  a  leg. 

3.  In  a  60°  triangle,  prove  that  the  hypotenusejs  2  times  as  long 
as  the  shorter  leg.     Prove  that  the  longer  leg  is  V3  times  as  long  as 
the  shorter  leg.     The  hypotenuse  is  how  many  times  as  long  as  the 
longer  leg? 

4.  Find,  correct  to  three  decimal  places,  V2;  V3;  V5. 

6.  If  the  hypotenuse  of  a  60°  triangle  is  two  units  in  length,  what 


§4] 


INTRODUCTION 


9 


is  the  length  of  the  shorter  leg  ?    Of  the  longer  leg  ?     If  the  hypotenuse 
is  four  units  in  length,  what  is  the  length  of  each  leg? 

6.  What  is  the  length  of  the  hypotenuse  of  a  45°  triangle,  if  each 


•C    5     w    ^ 


Ss   fc) 
3   ^ 


fc.  i:  ^ 

^        ^        T^ 


fe]  K| 


K 


leg   is  a  unit   in  length?     Two  units  in  length?     Three   units  in 
length? 

7.  What  is  the  length  of  each  leg  of  a  45°  triangle,  if  the  hypotenuse 
is  one  unit  in  length?     Two  units  in  length?     Three  units  in  length? 


10 


MATHEMATICS 


[§4 


8.  Find  the  length  in  rods  of  a  diagonal  of  a  section  of  land,  assum- 
ing that  the  section  is  a  true  square  1  mile  on  a  side. 

9.  A,  B,  C,  and  D  are  the  corners  of  a  section  of  land.     Through 
A,  a  line  is  drawn  making  an  angle  of  30°  with  the  side  AB,  and  inter- 
secting EC  at  the  point  E.     Find  the  length,  in  rods,  of  BE,  and  of 
AE.     Find  the  area  in  acres  of  the  triangular  plot  ABE. 

10.  Continuing  with  exercise  9,  let  F  be  the  mid-point  of  AB,  and 
G  the  mid-point  of  CD.  Draw  the  quarter  section  line  FG,  cutting 
the  line  AE  at  H.  Find  the  length  in  rods  of  AH,  and  of  FH.  Find 
the  area  in  acres  of  the  plot  AFH. 


FIG.  10. 


11.  The  longer  leg  of  a  plot  of  land  in  the  form  of  a  60°  triangle  is 
80  rods.     Find  the  area  of  the  plot  in  acres. 

12.  A  plot  of  land  in  the  form  of  a  60°  triangle  contains  72  acres. 
Find  the  length  in  rods  of  each  side  of  the  triangle. 

HINT:    Let  x  represent  the  number  of  rods  in  the  length  of  the 
shorter  leg. 

13.  The  shorter  side  of  a  rectangle  is  100  feet,  the  diagonal  is  200 
feet.     Find  the  length  of  the  longer  side. 


§4] 


INTRODUCTION 


11 


14.  The  diagonal  of  a  square  is  100  feet.  Find  the  length  of  a 
side. 

16.  A  barn  36  feet  wide  has  a  gambrel  roof.  The  lower  rafters 
make  an  angle  of  60°  with  the  horizontal.  The  upper  rafters  make  an 
angle  of  60°  with  the  vertical.  The  lower  and  upper  rafters  are  equal 
in  length.  Find  the  length  of  the  upper  edge  of  the  rafters.  See 
Fig.  10. 

HINT:    Let  x  be  the  number  of  feet  in  the  desired  length. 

16.  If  the  rafters  referred  to  in  exercise  15  are  2  X  6's,  and  if 
they  run  short  3/16  inch,  find  the  dimensions  of  the  block  sawed 
from  each  end.  (See  Fig.  11.)  Find  y  and  z. 


./A 


17.  If  the  building  referred  to  in  exercise   15  is  84  feet  long,  find 
the  number  of  cubic  yards 'space  under  the  roof  above  the  plates. 

18.  A  circular  concrete  silo,  12  feet  outside  diameter,  is  20  feet  tall. 
The  wall  is  8  inches  thick.     Find  the  number  of  cubic  yards  of  con- 
crete in  the  wall,  no  allowance  being  made  for  openings.     After  the 
problem  has  been  solved  by  the  student,  the  instructor  will  show 
how  algebraic  operations  may  be  used  in  eliminating  considerable 
numerical  work. 

19.  A  man  walks  3/4  mile    up  an  incline  making  an  angle  of 
30°  with  the  horizontal.     How  many  feet  does  he  move  in  a  hori- 
zontal direction? 


12  MATHEMATICS  [§5 

20.  A  ball  rolls  down  a  smooth  inclined  plane  making  an  angle 
of  45°  with  the  horizontal.  If  the  distance  s  measured  in  feet  along 
the  incline  is  given  by  the  formula 


where  t  is  time  in  seconds,  find  the  vertical  distance  through  which 
the  ball  moves  in  one  second;  in  two  seconds;  in  three  seconds. 

6.  Review  of  Elementary  Algebra.  The  course  for  which  this 
book  is  intended  presupposes  a  knowledge  of  elementary  algebra, 
and  of  plane  and  solid  geometry,  as  usually  covered  in  two  years 
of  High  School  work.  Many  students,  however,  having  had  their 
algebra  and  geometry  some  years  since,  may  feel  a  need  of  review 
in  these  subjects.  When  this  is  the  case,  the  student  should  take 
a  hasty  review  touching  only  the  essentials,  and  then  study  those 
subjects  or  theorems  which  he  wishes  to  apply,  at  the  time  he 
wishes  to  apply  them.  For  this  purpose,  a  very  brief  review  of 
algebra,  consisting  mostly  of  examples  and  illustrations,  is  here 
given. 

6.  Addition.  To  3x  +  4y  -  6z 
Add  2x  -  5y  +  2z 
Sum  5x  —  y  —  4z 

Exercises 

Find  the  sum  of  the  following: 

1.  3a  -  66  +  7c  and  5a  -  2c  -  76. 

2.  3x2  +  6xy  +  yz  and  x2  —  5xy  +  y*. 

3.  2xy  -  3x2y  +  3xy*  +  y2  and  2x2y  -  3y*  +  3xy  -  7. 

4.  a  +  b  +  c  -  d  and  2a  -  7c  -  2d  +  66  -  7. 

6.  a  -  26  +  6,     3a  +  7c  -  6  +  26,  and  5a  +  6  -  7c. 

6.  x2  +  2xy  +  y*,     x2  -  2xy  +  y2,  and  x2  -  2j/2. 

7.  3x3  +  6x2y  +  6xy2  +  y3,     5x3  —  Qy3  +  6x2y,     and     3x3   -  y3 


8.  a  +  b  +  c  +  3d,     2a  -  c  +  76  -  5d  +  6,  and  c  -  7  +  2d  -  a. 

9.  a3  -  63,     a26  +  a&2,  and  -  4o26  +  2a62. 

10.  25a3  -  2763,     5a26  -  6a62,  and  2663  -  24a3  +  a62. 

7.  Subtraction.  From  3a  -    66  +  7c  -  6 

Subtract          2a  +    56  -  2c  -  6 
Difference        a  —  116  +  9c 


§8]  INTRODUCTION  13 

Exercises 

1.  From  3o2  -  262  -  Sab  subtract  a2  -  3a6  +  62. 

2.  From  x2  -  2xy  +  y2  subtract  x2  +  2xy  +  y2. 

3.  From  x  +  2xz  +  3x3  -  2  subtract  2x  +  x2  +  7x3. 

4.  From  x3  +  3x2y  +  3xy2  +  y3  substract  Qx2y  +  2y3. 

5.  From  3x2  -  2y  +  y2  subtract  2x2  -  2y  +  6  +  7z2. 

8.  Multiplication.  The  product  of  two  numbers  having  like 
signs,  either  positive  or  negative,  is  positive.  The  product  of  two 
numbers  having  unlike  signs  is  negative.  Thus, 

(a)  (6)  =  ab 
(_a)(-6)  =  ab 
(a)(-  6)  =  -  ab 
(-a)  (6)  =  -ab 
Multiply       3z2  -   2xy  +  y2 
By  2x2  -     sy  +  y2 


-  2xt/3 


___  _ 
Product         6a;4  -  7x37/  +  7x2y2  -  3xy3  +  y* 


Exercises 

1.  Multiply  x3  +  x?/  +  y3  by  x  —  y. 

2.  Multiply  x3  —  xy  +  y3  by  x  +  y- 

3.  Multiply  x2  +  xy  +  y2  by  x2  -  xy  +  y2. 

4.  Multiply  x  —  7y  by  x  —  2y. 

5.  Multiply  x  +  y  by  x  —  y. 

6.  Multiply  3x  +  7y  by  3x  +  8y. 

7.  Multiply  a  +  b+c+dbya  +  b-c-d. 

8.  Multiply  2x  -  3y  +  6  by  2x  +  7y  +  2. 

9.  Multiply  3xy  -  2x2  -  2x  by  2xy  -  2x  -  3y  -  6. 
10.  Multiply  9  -  3x  +  x2  by  3  +  x. 

A  few  simple  multiplications  may  be  performed  mentally.  We 
see  at  once  that  (a  +  6)  (a  —  6)  =  a2  —  62,  where  a  and  b  are  any 
two  numbers.  From  this  we  have 

(3x  -  2y)(3x  +  2y)  =  9x2  -  4?/2 


14  MATHEMATICS 

Exercises 

Multiply  the  following  mentally: 

1.  (3z  -y)(3x+y). 

2.  (2x  -y)(2x  +  y). 

3.  (a  -  9)  (a  +  9). 

4.  (x*y  -3a)(x2?/  +  3o). 
6.  (a  +36)  (a  -  36). 

6.  (29)  (31),  or  (30  -  1)(30  +  1)  =  900  -  1  =  899. 

7.  (51)(49). 

8.  (52)  (48),  or  (50  +  2)  (50  -  2). 

9.  (103)  (97). 
10.  (25)  (35). 

A  few  powers  of  binomials  are: 

(a  +  6)2  =  a2  +  2ab  +  b\ 

(a  -  6)2  =  a2  -  2ab  +  62. 

(a  +  6)>  =  a3  +  3o26  +  3a62  +  b3  . 

(a  -  6)3  =  a3  -  3a26  +  3a&2  -  b3. 

(a  +  &)4  =  a4  +  4a36  +  6a262  +  4a63  +  64. 

(a  -  6)4  =  a4  -  4a36  +  6a262  -  4a63        4 
Thus,  (3  -  a)3  =  27  -  27a  +  9a2 

and  (x  +  ?/2)4  =  x4 


Exercises 

Expand  the  following  mentally: 

1.  (2o  -  a;)2.  4.  (x  -  3)4. 

2.  (x  +  3y)2.  5.  (1  -  z)3. 

3.  (2z  -  I)3.  6.  (2  +  ?/)4. 

7.  (52)2,  or  (50  +  2)2,  or  2500  +  200  +  4  =  2704. 

8.  (31)2,  or  (30  +  I)2. 

9.  (29)2,  or  (30  -  I)2. 
10.  (98)2. 

The  square  of  a  polynomial  is  illustrated  by  the  following: 

(a  +  6  +  c)2  =  a2  +  b2  +  c2  +  2ab  +  2ac  +  2bc. 
(a  +  6  +  c  +  d)2  =  a2  +  62  +  c2  +  rf2  +  2o6  +  2ac  +  2 

+  2bc  +  26rf  +  2cd. 
(3  -  x  +  ?/2)2  =  9  +  xz  +  y*  -  6x  +  6i/2  -  2^. 


§9]  INTRODUCTION  15 

Exercises 
Expand  the  following  mentally: 

1.  (o  +  b  +  2)2.  4.  (2o  -  x  +  3)2. 

2.  (a  +  b  -  2)2.  6.  (x2  -  2?/2  +  4)2. 

3.  (a  -  b  -  c)2. 

The  product  of  two  binomials  having  a  common  term  may  be 
found  by  inspection.  Illustrations: 

(a  +  6)  (a  +  c)  =  a2  +  (b  +  c)o  +  be. 
(x  +  7)(x  +  2)  =  z2  +  9z  +  14. 
(x  -  5)(z  +  3)  =  x2  -  2x  -  15. 
(x2  -  2y)(x*  -  By)  =  x*  -  5x2y  +  6?/2. 

Exercises 

Find  mentally  the  value  of  each  of  the  following: 

1.  (x  +  2)(x  +  3).  6.  (3x  +  2y)(3x  -  7y). 

2.  (x  -  2)(x  +  3).  7.  (x2  -  3)(x2  -  4). 

3.  (x  -  2)(x  -  3).  8.  (3xy  -  z)(3xy  +  7z). 

4.  (x  +  2)(x  -  3).  9.  (xV  -  3)(xV  -  10). 

5.  (x2  +  5y)(x2  -  5y).  10.  (x  -  2y)(2x  -  2y). 

9.  Division.  The  quotient  of  two  numbers  having  like  signs 
is  positive.  The  quotient  of  two  numbers  having  unlike  signs  is 
negative. 

Thus,  (a)  -(&)=- 


f    \     •     f         M  a 

(0)  T  (  _  6)  =  _  _ 

(-a)-*-  (6)=  '-£ 

Divide  a;3  -  Go;2  -  19x  +  84  by  x  -  7 
x  -  7(x*  -  6z2  -  19x  +  84)  z2  +  a;  -  12     Quotient 


x2  -19z  +  84 

a2  -  7x 
-  12x  +  84 
-12x+  84 


16  MATHEMATICS  [§10 

Exercises 

1.  Divide  15x2  -  llx  -  14  by  3x  +  2. 

2.  Divide  12a2  -  28a  +  15  by  6a  -  5. 

3.  Divide  8m5  -  14m2  -  18m  +  21  by  4m3  +  6m  -  7. 

4.  Divide  3x5  -  7x3  -  2x  -  36  by  x  -  2. 

5.  Divide  x8  +  2x6  -  3x4  +  7x3  -  2x  +  69  by  x  -  3. 

6.  Divide  x4  +  x2y2  +  y*  by  x2  +  xy  +  y2. 

7.  Divide  8x4  -  22x3y  +  43x2t/2  -  38xy3  +  24y*  by 
2x2  -  3xy  +  4y2. 

8.  Divide  x3  —  y3  by  x  —  y. 

9.  Divide  x4  —  y4  by  x  +  y. 
10.  Divide  x4  —  y4  by  x  —  y. 

10.  Symbols  of  Aggregation.     If  a  sign  of  aggregation  is  pre- 
ceded by  the  negative  sign,  change  all  signs  within  when  the  sign 
of  aggregation  is  removed.    If  the  sign  of  aggregation  is  preceded 
by  the  positive  sign,  all  signs  within  remain  the  same  when  the 
sign  of  aggregation  is  removed. 

5x2  -  [3y2  +  {2x2  -  (?/2  +  3x2)  +  5y*}  -  x*] 

=  5x2  -  [3y2  +  {2x2  -  y2  -  3x2  +  5r/2}  -  x2] 

=  5x*  -  [3y2+  {4t/2  -x"'}  -  x2] 

=  5x2  -  [3y2  +  4y2  -  xz  -  x2] 

=  5x2  -  [7y2  -  2x2] 

=  5x2  -  7y2  +  2a;2 

=  7x2  -  7y2 

Exercises 

Simplify  the  following  by  removing  the  signs  of  aggregation : 

1.  ab  -  462  -  (2a2  -  62)  -  {   -  5a2  +  2ab  -  362} . 

2.  x  -  { y  +  z  -  [x  -  (  -  x  -  y)  +  z}}   +  [z  -  (2x  -  y)]. 

3.  a  -  {   -  a  -  [  -  a  -  (  -a  -  1)]}. 

4.  3yz  -  [2yz  +  (9z  -  2yz}]. 

5.  -  {   -  1  -[-1  -  (  -1)]}. 

11.  Factoring.     Since   (a  +  6)2  =  a2  ±  2ab  +  b2,   any   expres- 
sion of  the  form  of  the  right-hand  side  can  be  factored  by  inspec- 
tion.    Thus, 

x2  -  6xy  +  Qy2  =  (x  -  3y)2 
and 

:.„      4  +  4(o  +  6)  +  (a  +  6)2  -  (2  +  a  +  6)2 


§11]  INTRODUCTION  17 

Exercises 

Factor  the  following  by  inspection : 

1.  9x2  -  30xy  +  25y2. 

2.  4  +  IQt  +  16<2. 

3.  zV  +  lOxVz2  +  25z4. 

4.  9  +  6(x2  +  t/2)  +  (x2  +  y2)2. 

5.  a4  +  4a262  +  464. 

Since  (a  +  6)  (a  —  6)  =  a2  —  62,  any  expression  of  the  form  of 
the  right-hand  side  can  be  factored  by  inspection.    Thus, 
4a2  -  962  =  (2a  +  36)  (2o  -  36) 

Exercises 

Factor  the  following  by  inspection : 

1.  xV  -  22.  4.  25  -  3x2. 

2.  (a  +  6)2  -  c2.  5.  81  -  625x4. 

3.  c2  -  (a  +  6)2. 

Since  (a  +  6)  (a  +  c)  =  a2  +  (6  +  c)a  +  &c,  any  expression  of 
the  form  of  the  right-hand  side  can  be  factored  by  inspection. 
Thus, 

x2  -  5x  -  14  =  (x  -  7)(x  +  2) 

Exercises 

Factor  the  following  by  inspection : 

1.  x2  +  7x  +  10.  4.  9x2  -  18x  -  27. 

2.  a2  +  4ay  -  21y2.  5.  25  +  30a  -  27a2. 

3.  4x2  -  ISxy  +  18j/2. 

Since  (a  +  6)  (a2  —  ab  -f-  62)  =  a3  +  &3,  any  expression  of  the 
form  of  the  right-hand  side  can  be  factored  by  inspection.    Thus, 
27  +  125x3  =  (3  +  5x)(9  -  I5x  +  25z2) 

Exercises 

Factor  the  following  by  inspection: 

1.  zV  +  1-  4-  125  +  xy. 

2.  x6  +  y\  6.  x3  +  8y6. 

3.  8  +  27x3. 

Since  (a  —  6)  (a2  +  06  +  62)  =  a3  —  b3,  any  expression  of  the 
form  of  the  right-hand  side  can  be  factored  by  inspection.    Thus, 
27  -  125z3  =  (3  -  5s)  (9  +  15*  +  25z2) 


18  MATHEMATICS  [§12 

Exercises 

Factor  the  following  by  inspection : 

1.  x3y3  -  1.  4.  125  -  x3y3. 

2.  x6  -  y6,  or  (x3  +  y3)(x3  -  y3).  5.  x3  -  8y*. 

3.  8  -  9x3. 

The  following  may  be  factored  by  grouping  the  terms.     Thus, 
a3m  +  a3n  —  m  —  n 

=  a3(m  -\-  n)  —  (m  -\-  n) 

=  (a3  -  l)(m  +  n) 

=  (a  -  l)(a2  +  a  +  l)(m  +  n) 

Exercises 

Factor  the  following : 

1.  ax  —  ay  +  bx  —  by.  4.  x5  —  xy*  —  x*y  +  yb. 

2.  x3  +  3x2  +  3x  +  1.  6.  x4  -  xsy  -  xy3  +  y*. 

3.  ax2  -  2axy  +  ay2  +  6x2  -  2bxy  +  by*. 

12.  Fractions.  Multiplying  or  dividing  both  numerator  and 
denominator  of  a  fraction  by  the  same  number,  excepting  zero, 
does  not  change  the  value  of  the  fraction.  To  reduce  a  fraction 
to  its  lowest  terms,  factor  both  numerator  and  denominator  and 
then  divide  each  by  the  common  factors  if  there  are  any.  Thus, 
ax*  +  ax2y2  +  ay* 

a2z3  —  a2y3 
a(x*  +  xy  +  t/2)(a;2  -  xy  +  yz] 


a(x  —  y)(x2  -\-xy-\-  y2) 
x2  —  xy  +  y2 
(x  -  y) 

Exercises 

Reduce  the  following  to  lower  terms: 
^     ox2  —  ay2  >  ax  +  ay  —  x  —  y. 

x3  —  y3  '        x3  +  y3 

x2  +  x  -  6_  27  -  x3 

2*        x2  -  4  12-  7x  +  x2. 

•    x4  -  y\ 

x6-6?/ 


§12]  INTRODUCTION  19 

To  multiply  fractions  together,  multiply  the  numerators 
together  for  the  numerator  of  the  product,  and  multiply  the 
denominators  together  for  the  denominator  of  the  product.  To 
simplify  the  result,  divide  both  numerator  and  denominator  by 
common  factors  if  there  are  any.  Thus, 

a3  -  b3  b  a26  -  ab2 

ab      Xa2  +  ab  +¥  X  (a  -b)z 
(a3  -  63)(6)(a26  -  a&2) 
"  ab(a2  +  ab  +  62)(a  -  by 
_  (a  -  6)  (a2  +  ab  +  62)(a62)(a  -  6) 

a&(a2  +  ab  +  62)(a  -  6)2 
=  6 

In  working  this  problem,  the  second  expression  should  not  be 
written  down,  and  possibly  the  third  could  be  omitted. 

Exercises 

Simplify  the  following : 
a4-  y* 


a2  -  2ay  +  y2 
az(x  +  a) 
x2  +  lax  +  a 


a2  +  2a.b  +  b2  -  4c2        (a  -  b)2  -  4c"        (2a  -  b)2  -  4c2. 
4a2  -  (b  +  2c)2  (a  +  6)'  -  4c2        a2  -  (b  -  2c)2 

x3  -  y3       x*  —  xy  +  y*       x  +  y% 
s\    .    ,  ,  X 


•7*3    _L_   4/3  /7«2    _!      TII    —1—   -i/2  />•    _ _    91 

•*'     \  y        *    \  -Ly   i  y        *      j/ 
_    oa;  +  ay      bx  —  36       x2  —  xy  +  y2 

b         <  x»+ j/3  X  x2  -  5x  +~6~' 

To  add  fractions  reduce  each  to  the  same  denominator  by  multi- 
plying numerator  and  denominator  of  each  by  a  suitable  number, 
then  add  the  numerators  and  place  the  sum  over  the  common 
denominator.  Thus, 

x  -y  _J_  xy 

I       -v      I      ..      I       _3       I 


x2  -  zt/  +  7/2   '    a:  +  y   '   x3  +  y3 
x2  -  y*      x2  -  xy  +  y2          xy 
x3  +  y3  "*        a;3  +  y3       +a;3  +  y3 
2a:2 


20  MATHEMATICS  [§13 

Exercises 

Simplify  the  following : 

•  •     i       n      I  o      I      o  f* 


2      i  v-2 


+  3). 


a;  —  5       x  —  1 


.    o 


x-  5      x»  -  te  +  5 
y)       («  +  y)*' 


y          (*-»)« 

13.  Exponents, 

Definitions:     an  =  a  a  a  .  .  .  to  n  factors,  if  n  is  a  positive 
integer 

ap/«  =  v/ap,  if  p  and  g  are  positive  integers 
a«  =  1 
a~8  =  -,  if  s  is  positive 

ft8 

Laws  of  indices:    aman  =  am+" 


(am)«  =  amn 


Exercises 

1.  Express  the  following  using  positive  exponents  only: 
(a)  x~2ab~3.  ar^-'x-*. 

(6)  a-^b^.  W       c2«Sa;6 

a~lbc~* 
(c)    (x-^y-^)«.  (e)    ^7c^i- 

2.  Perform  the  indicated  operations: 

(a)  25W.  (6)  8?4.  (c)  (9/16)-*.  (d)  27°. 

3.  Simplify: 


§14]  INTRODUCTION  21 

4.  Multiply  Sa%  -  4o6w  +  2ow6  -  &?*  by  2ow  -  36^. 
6.  Divide  a*  -  3a2  +  a^  -  4  +  12ow  -  4o%  by  a**  -  4. 

14.  Radicals. 

Simplify     ^24  +  5^81  -  6^192  -  10^1029 

+ 


=  2  i/3  +  15^3  -  24i/3  -  70^3 
=  -  77^/3 

Rationalize  the  denominator  of 

(\/2  +  \/3)(>/2  +  \/3) 


\/2  -  \/3        (\/2  - 


2-3 

=  2  +  2\/6  +3 
-  1 

=  -  (5  +  2V6) 

Exercises 

1.  Simplify  Vl2  -  2\/27  +  9 A/48  -  Vl08  -  6\/75. 

2.  Simplify  V (a  -  b)*x  -  V(o  +  6)2z  -  Va2x  + 

2 

3.  Rationalize  the  denominator  of 


4.  Rationalize  the  denominator  of 


V3  -  V2 

2  -  \/5 

6.  Rationalize  the  denominator  of  —7= 7=' 

V3   -  V  5 

16.  Quadratic  Equations.  A  quadratic  equation  may  be  solved 
by  completing  the  square,  or  by  using  the  formula,  or  somtimes 
by  factoring. 

Solve  for  x:    x2  -  6x  +  8  =  0. 
By  completing  the  square  we  have 

x2  -  6z  +  9  =  1, 
or  x  -  3  =  ±  1, 

or  x  =  3  +  1, 

or  x  =  4  or  2. 


22  MATHEMATICS  [§15 

By  factoring  the  left-hand  side  we  have 

(x  -  4)(x  -  2)  =  0. 

This  equation  is  equivalent  to  the  two  equations. 

x  -  4  =  0 

and  z  -  2  =  0 

from  which  we  have 

x  =  4 
and  x  =  2 

If  axz  +  bx  +  c  =  0  is  the  general  equation  of  the  second 
degree,  its  solutions  are 

-  b  +  V'b2  -  4ac 

2a 

In  the  above  equation,  a  =  1,  b  =  —  6,  andc  =  8.     Substitut- 
ing these  values  in  the  formula  we  have 

6  +  V31T-4-8 
x  —  — '• 


2. 

=  3  ±  1, 
or  x  =  4  or  2 

Exercises 

Solve  the  following  equations  for  x: 

1.  x2  -  9x  +  20  =  0. 

2.  4z2  +  7x  -  2  =  0. 

3.  x2  -  7z  +  6  =  0. 

4.  8x2  -  4x  +  3  =  0. 
6.  9x2  +  7x  -  20  =  0. 


CHAPTER  I 
GRAPHIC  REPRESENTATION 

The  Straight  Line.     The  Parabola.     The  Equilateral  Hyperbola. 

16.  The  Scale.     In  solving  problems  graphically  the  funda- 
mental instrument  used-  is  the  scale.     Unless  otherwise  stated, 
the  term  scale,  as  used  in  this  book,  means  what  is  commonly 
understood  by  a  scale,  an  instrument  for  measuring  short  linear 
distances.     The  term  scale,  in  a  broader  sense,  may  be  defined  as 
any  curve,  including  a  straight  line,  the  points  along  which  are 
supposed  to  correspond  in  order  to  the  numbers  of  a  sequence. 
We  are  all   familiar  with  various  scales,  such  as  the  foot-rule, 
divided  into  inches;  the  meter,  divided  into  centimeters  and  milli- 
meters; the  thermometer  scale,  divided  to  correspond  to  degrees 
and  tenths  of  degrees  of  temperature;  the  scale  on  a  burette, 
divided  to  correspond  to  cubic  centimeters  and  tenths  of  cubic 
centimeters;  and  scores  of  other  scales  used  for  measuring  different 
quantities.     If  a  scale  is  marked  so  that  the  subdivisions  are  all 
of  equal  length,  as  the  foot-rule,  we  say  the  scale  is  divided  uni- 
formly, or  that  we  have  a  uniform  scale;  otherwise,  the  scale  is  a 
non-uniform  scale.     For  the  solution  of  problems  graphically,  the 
student  should  have  for  measuring  lengths  a  scale  divided  into 
centimeters  and  millimeters  or  into  inches  and  tenths  of  inches. 
Illustrations  of  non-uniform  scales  will  be  given  later. 

17.  Drawing  to  Scale.     A  large  number  of  graphic  problems 
consist  in  nothing  more  than  drawing  to  scale,  i.e.,  making  a 
picture  of  some  object  in  which  a  definite  length,  as  the  inch,  on 
the  drawing  represents  a  definite  length,  as  the  foot,  on  the  given 
object.     If  the  dimensions  on  such  a  drawing  were  to  be  given  in 
feet  and  inches,  or  if  dimensions  in  feet  and  inches  were  to  be 
scaled  from  the  drawing,  a  scale  divided  into  tenths  of  inches 
would  apparently  be  very  ill-adapted.     For  this  kind  of  work, 
scales    may  be   procured  on  which  the  unit  of   length  is  sub- 

23 


24 


MATHEMATICS 


[§18 


divided  into  twelve  equal  parts.    If  each  unit  throughout  the 
entire  length  of  the  scale  is  subdivided,  the  scale  is  called  a  full 

_  divided    or   chain  scale.      If 

only  one  unit,  usually  the 
left-hand  unit,  is  subdivided, 
it  is  called  an  open  divided 
scale.  These  scales  are  made 
with  the  unit  J,  \,  f,  £,  \,  1, 
2,  2,  3,  etc.,  inches  long,  used, 
respectively,  when  |,  j,  f ,  |,  $, 
1,  2,  2,  3,  etc.,  inches  on  the 
drawing  are  to  represent  1 
foot  on  the  object. 

Fig.    12   represents   an  18- 
^      inch,  flat,  open  divided  scale. 
%         The  unit   on  one  edge  is  f 
^      inch,  and  on  the  other  edge 
]>      it  is  j  inch.     This  scale  may 
^      be  used  in  drawing  the  plans 
g,     of  a  barn.     The  small  scale 
9      is  used  in  drawing  the  plan 
^      and  elevations  ?  inch  to  the 
foot,  while  the  larger  scale  is 
2      used  in  drawing,  on  a  larger 
scale,  the  details  of  the  plank 
frame    truss    bents.         The 
method  of  measuring  lines  is 
illustrated  in  the  figure.     The 
line  AB  is  found  to  be  22  feet, 
7  J  inches  long. 

18.  Positive  and  Negative 
Directions.  If  upon  a  scale 
some  point  is  chosen  as  zero, 
or  origin,  distances  measured 
in  one  direction,  or  the  num- 
bers represented  by  the  points 
on  one  side  of  it,  are  called  positive,  and  are  designated  with  the 
positive,  or  plus,  sign  +.  Distances  measured  in  the  other 


§19] 


GRAPHIC  REPRESENTATION 


25 


direction,  or  numbers  represented  by  points  on  the  other  side  of 
the  origin,  are  called  negative,  and  are  designated  by  the  nega- 
tive, or  minus,  sign  — . 

19.  Rectangular  coordinate  paper,  or  squared  paper,  is  paper 
ruled  in  both  directions,  horizontally  and  vertically.  These 
rulings  are  generally  spaced  uniformly,  dividing  the  page  into  a 
number  of  equal  squares  or  rectangles.  The  size  of  these  squares 
or  rectangles  is  entirely  arbitrary.  The  length  of  a  side  may  be 
an  inch,  a  tenth  of  an  inch,  a  centimeter,  a  millimeter,  etc., 


x- 


Y' 
FIG.  13. — Coordinate,  or  squared  paper. 

depending,  to  some  extent,  upon  the  use  for  which  the  paper  is 
designed.  For  convenience,  every  fifth,  tenth,  fourth,  eighth,  or 
twelfth  line  is  made  somewhat  heavier  than  the  others.  A  sheet 
of  squared  paper  is  represented  in  Fig.  13.  Upon  this  sheet  two 
lines,  mutually  at  right  angles,  are  chosen  as  axes.  The  horizontal, 
X'OX,  is  called  the  axis  of  abscissas;  the  vertical,  Y'OY,  the 
axis  of  ordinates.  The  point  of  intersection,  0,  of  these  axes  is 
called  the  origin. 

A  use  of  squared  paper  is  illustrated  by  the  following:  In 
Table  I  are  given  the  maximum  and  minimum  air  temperatures 
at  Madison,  Wisconsin,  for  each  day  of  the  month  of  October,  1910. 


26 


MATHEMATICS 


[§19 


TABLE   I.— MAXIMUM  AND  MINIMUM  TEMPERATURES  AT  MADISON 
WISCONSIN,  FOR  OCTOBER,  1910 


Date 

Max.  temp., 
°F. 

Min.  temp., 
°F. 

Date 

Max.  temp., 
°F. 

Min.  temp. 

op 

1 

66 

55 

17 

81 

53 

2 

71 

48 

18 

81 

57 

3 

75 

58 

19 

69 

45 

4 

68 

55 

20 

45 

40 

5 

62 

53 

21 

49 

41 

6 

58 

45 

22 

52 

34 

7 

66 

43 

23 

60 

37 

8 

68 

47 

24 

60 

49 

9 

60 

44 

25 

52 

44 

10 

67 

42 

26 

60 

40 

11 

75 

49 

27 

42 

32 

12 

61 

46 

28 

35 

30 

13 

69 

45 

29 

38 

26 

14 

73 

52 

30 

60 

31 

15 

76 

50 

31 

63 

39 

16 

80 

56 

These  data  are  said  to  be  represented  graphically  in  Fig.  14,  or  the 
data  are  said  to  be  plotted  upon  squared  paper.  To  do  this:  first, 
points  equidistant  along  the  axis  of  abscissas  are  chosen  to  repre- 
sent the  days  of  the  month.  These  points  are  numbered  con- 
secutively from  1  to  3 1.1  Distances  from  this  horizontal  line 
represent  temperatures.  If  the  temperature  is  above  zero,  or 
(+),  it  is  represented  by  a  distance  above  the  horizontal  axis; 
if  below  zero,  or  (  — ),  by  a  distance  below.  The  distance  above 
or  below  is  proportional  to  the  magnitude  of  the  temperature 
reading.  Since  distances  from  the  horizontal  axis  represent  tem- 
perature readings,  points  along  the  axis  of  ordinates  are  numbered 
to  correspond  to  degrees  of  temperature.  In  the  figure,  the  verti- 
cal distance  between  two  adjacent  horizontal  lines  represents 
2°,  but  only  every  fifth  line  is  'actually  numbered,  since 
numbering  each  line  would  not  aid  in  any  way  in  reading  the 
drawing.  This  vertical  axis  is  in  reality  a  uniform  scale,  for  equal 

1  Only  the  even  numbers  are  placed  in  the  drawing;  otherwise  the  numbers  would 
be  so  near  together  as  to  be  confusing.  Omitting  the  numbers  from  every  other 
point  does  not  make  the  drawing  harder  to  read. 


§19] 


GRAPHIC  REPRESENTATION 


27 


-^ 


<^ 


^ 


\D 


V 


v 


1 


<<; 


> 


> 


> 


H 

3 
p 

<j 

CS 

? 

PW       ^^ 

S     « 
H     % 


o  S 

P     Q  ° 


S 

o 

O        TJH 


28  MATHEMATICS  [§20 

portions  of  it  represent  equal  changes  in  temperature.  Since  the 
scale  is  uniform,  it  is  evident  that  it  is  immaterial  whether  we 
think  of  the  point  or  the  vertical  distance  from  the  origin  to  that 
point  as  representing  degrees  of  temperature.  Further,  we  may 
look  upon  the  scale  as  continuous,  for  every  point  upon  it  cor- 
responds to  some  temperature. 

The  horizontal  axis,  as  we  are  here  using  it,  is  not  a  continuous 
scale  for  measuring  time,  for  each  numbered  point  along  it,  or 
the  distance  of  this  point  from  the  origin,  represents  a  whole 
interval  of  time  from  midnight  to  midnight;  and  the  intermediate 
points  have  no  meaning  whatever. 

Now  that  we  have  chosen  points  to  represent  the  days  of  the 
month,  a  unit  of  length  to  represent  a  degree  of  temperature,  and 
have  marked  the  axes,  we  are  ready  to  transfer  the  readings  from 
the  table  to  the  squared  paper.  On  the  first  of  October,  the  maxi- 
mum temperature  was  +  66°.  Graphically,  this  is  represented  by 
making  a  dot  or  small  circle  thirty-three  divisions  above  the  hori- 
zontal axis  and  upon  the  vertical  line  passing  through  that  point 
of  the  horizontal  axis  representing  the  first  day  of  October.  This 
reading  is  now  said  to  be  plotted.  The  maximum  temperature 
on  the  second  of  October  was  71°,  and  is  represented  by  placing 
a  point  or  circle  thirty-five  and  one-half  spaces  above  the  hori- 
zontal axis  and  upon  a  vertical  line  passing  through  that  point 
upon  the  horizontal  axis  marked  2.  In  a  similar  way,  points 
representing  the  maximum  and  minimum  temperatures  for  each 
day  are  plotted.  To  aid  the  eye  in  following  the  temperatures 
from  day  to  day  the  consecutive  points  are  connected  by  short 
straight  lines.  These  broken  lines  will  also  aid  one  to  see  at  a 
glance  how  the  maximum  and  minimum  temperatures  rose  and 
fell  from  day  to  day  throughout  the  month;  to  see  when  the 
temperature  was  high;  when  low;  when  the  freezing  point  was 
reached;  etc.  In  short,  Fig.  14  is  a  picture  of  all  that  is  given 
in  Table  I. 

20.  Smooth  Curves  Drawn  Through  Plotted  Points.  In  Fig. 
15  are  plotted  the  data  given  in  Table  II,  the  hourly  air  tempera- 
tures for  May  14  and  October  10,  1910,  at  Madison,  Wisconsin. 
It  will  be  noticed  that  in  this  drawing  the  points  are  connected, 
not  by  straight  lines  as  in  the  previous  illustration,  but  by  means 


§20] 


GRAPHIC  REPRESENTATION 


29 


of  a  smooth  curve  drawn  through  the  points.  The  reason  for 
drawing  a  smooth  curve  is  that  the  readings  were  taken  so  near 
together  that  the  points  along  the  smooth  curve  are  probably  very 
near  the  points  that  would  have  been  obtained  had  additional 
intermediate  temperature  readings  been  taken.  Thus  the  curves 
in  Fig.  15  may  be  used  for  scaling  off  temperatures  for  any  time  of 
the  day. 


•'£  60 

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— 

— 

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Hourly  Air 

Te 

at 
n  V 

r,  l 

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t..  I 

•nperatures 

Hm 

Ma< 

liso 
Ma 

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Wisconsin 
4.1910 
i 
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2          4          6         8          10        12^       2          4        ^6          8         10        1 

<                           A.M;                         >  |  <                           F.M.                          * 
Time  of  Day  Hours 

FIG.  15. 

TABLE  II.— HOURLY  AIR  TEMPERATURES  AT  MADISON,  WISCONSIN. 
MAY  14,   1910;  OCTOBER,  10,  1910 


Time 

May  14,  1910, 
temp.,  °  F. 

Oct.  10,  1910, 
temp.,  °  F. 

Time 

May  14, 
1910, 
temp.,  °  F. 

Oct.  10, 
1910, 
Temp.,  °  F 

1  a.  m. 

37 

44 

1  p.  m. 

58 

65 

2  a.  m. 

35 

44 

2  p.  m. 

61 

66 

3  a.  m. 

35 

44 

3  p.  m. 

63 

67 

4  a.  m. 

34 

43 

4  p.  m. 

62 

67 

5  a.  m. 

34 

43 

5  p.  m. 

62 

65 

6  a.  m. 

35 

42 

6  p.  m. 

61 

62 

7  a.  m. 

37 

43 

7  p.  m. 

59 

57 

8  a.  m. 

42 

45 

8  p.  m. 

56 

55 

9  a.  m. 

46 

51 

9  p.  m. 

53 

55 

10  a.  m. 

49 

55 

10  p.  m. 

50 

55 

11  a.  m. 

54 

60 

llp.m. 

47 

52 

12  a.  m. 

56    !    62     12p.m. 

45 

51 

30 


MATHEMATICS 


[§21 


21.  Precipitation  Charts.  In  Fig.  16  we  have  represented 
diagrammatically,  by  the  usual  method,  the  mean  monthly  pre- 
cipitation at  Tacoma,  Washington;  Madison,  Wisconsin;  Tampa, 
Florida;  and  Yuma,  Arizona.  In  these  charts  the  amount  of  pre- 
cipitation is  represented  by  the  height  of  a  shaded  rectangle.  This 
method  of  representation  is  similar  to  that  used  in  Fig.  14  and 
enables  one  to  see  at  a  glance  when  the  rainy  and  the  dry  periods 
occur  at  these  localities. 


J  FMAM J  JASOND 
Tacoma,  Wash. 


J  FMAM  J  JASOND 
Tampa,  Fla. 


.  6- 
a> 

-S5 

a  4. 

13 

.2  3 
3  2- 


TIM          Ml 
I  I  I  I  I  I  I  I  IN 


J  F MAM J  J  A  S  O  N  D 

Madison,  Wis. 


P-iU 


J  FMAM  J  JASOND 
Yuma,  Ariz. 


FIG.  16. — Precipitation  charts. 

22.  Marking  Axes.  Whenever  data  are  represented  upon 
squared  paper,  both  horizontal  and  vertical  axes  should  be  marked 
in  such  a  way  as  to  make  perfectly  clear  what  quantities  are 
represented  and  what  length  represents  a  unit  of  measure  of  that 
quantity.  If  two  or  more  curves  are  drawn  upon  the  same  set  of 
axes,  they  should  be  so  marked  as  to  show  definitely  what  each 
curve  represents. 

Occasions  may  arise  when  it  is  desirable  to  draw  upon  the  same 


§22]  GRAPHIC  REPRESENTATION  31 

sheet  of  paper,  for  comparison,  two  curves  having  in  common  only 
one  element,  such  as  time.  A  scheme  for  doing  this  is  illustrated 
by  exercise  12,  below. 

The  length  chosen  to  represent  a  unit  is  entirely  arbitrary  and 
need  not  be  the  same  along  both  axes.  This  is  already  illustrated 
in  Fig.  15.  Further,  the  numbering  in  the  horizontal  and 
vertical  directions  need  not  be  placed  upon  the  axes,  as  will  be 
illustrated  by  exercise  12,  below. 

Exercises 

1.  At  what  hour  was  the  temperature  a  maximum  on  May  14, 
1910?     At  what  hour  was  the  temperature  a  maximum  on  Oct.  10, 
1910? 

2.  At  what  time  was  the  temperature  a  minimum  on  May  14, 
1910?     At  what  time  was  the  temperature  a  minimum  on  Oct.  10, 
1910? 

3.  At  what  hour  was  the  temperature  increasing  the  fastest? 
At  what  hour  was  the  temperature  decreasing  the  fastest? 

4.  During  what  hours  of  the  day  was  the  temperature  decreasing? 
During  what  hours  of  the  day  was  the  temperature  increasing? 

5.  What  was  the  temperature  at  10:30  a.  m.?     10:30  p.  m.? 

6.  What  was  the  temperature  at  6: 15  a.  m.?     6: 15  p.  m.? 

7.  At  what  time  was  the  temperature  on  May  14  the  same  as  on 
October  10? 

8.  For  what  portion  of  the  day  was  it  warmer  on  May  14  than  on 
October  10? 

9.  During  what  hours  was  the  temperature  below  the  freezing 
point? 

10.  Plot  data  given  in  Table  III. 

11.  Plot  data  given  in  Table  IV. 

12.  Plot  the  data  given  in  Table  V  upon  squared  paper.     Use  the, 
same  horizontal  axis  for  all  three   curves.     Put  the  temperature 
curves  above  the  discharge  curve,  using  the  same  horizontal  (time) 
scale   for   both.     Let    1   cm.    (or   1/2  inch)    on  the   vertical    scale 
represent  0.1    second-foot1  discharge,  and   10°   temperature.     Start 
the   temperature   scale   with    60°,   and   place   the  60,  6  cm.   (or  3 
inches)  above  the  horizontal  scale.     Start  the  discharge  scale  with 
3.4  placed  on  the  horizontal  scale. 

1  Second-foot  (or  sec-ft.),  when  applied  to  the  measurement  of  flow  of  water 
means  one  cubic  foot  per  second. 


32 


MATHEMATICS 


[§22 


TABLE  III— MAXIMUM  AND  MINIMUM  TEMPERATURES  FOR 
OCTOBER,  1911,  AT  MADISON,  WISCONSIN 


Date 

Max.  temp., 
°F. 

Min.  temp., 
°F. 

Date 

Max.  temp., 
°F. 

Min.  temp., 
°F. 

1 

55 

50 

17 

61 

51 

2 

56 

50 

18 

67 

47 

3 

72 

50 

19 

57 

45 

4 

61 

48 

20 

54 

39 

5 

51 

42 

21 

46 

38 

6 

63 

47 

22 

48 

35 

7 

53 

45 

23 

42 

34 

8 

61 

40 

24 

46 

30 

9 

62 

45 

25 

47 

37 

10 

61 

49 

26 

38 

30 

11 

65 

48 

27 

42 

28 

12 

69 

49 

28 

43 

29 

13 

57 

47 

29 

53 

30 

14 

61 

50 

30 

45 

37 

15 

57 

49 

31 

40 

32 

16 

65 

52 

TABLE    IV.— HOURLY    TEMPERATURES    AT    MADISON,     WISCONSIN, 
FOR  APRIL  11,  1911,  AND  NOVEMBER  10,  1911 


Hour 

Temp.,  °  F., 
Nov.  10, 
1911 

Temp.,  °  F., 
April  11, 
1911 

Hour 

Temp.,  °  F., 
Nov.  10, 
1911 

Temp.,  °  F., 
April  11, 
1911 

a.  m. 
1:00 

59 

39 

p.  m. 
1:00 

62 

59 

2:00 

59 

40 

2:00 

60 

61 

3:00 

59 

40 

3:00 

59 

59 

4:00 

59 

40 

4:00 

51 

58 

5:00 

60 

40 

5:00 

44 

56 

6:00 

60 

40 

6:00 

39 

53 

7:00 

61 

41 

7:00 

35 

53 

8:00 

62 

44 

8:00 

31 

52 

9:00 

64 

47 

9:00 

28 

51 

10:00 

66 

50 

10:00 

25 

47 

11:00 

68 

55 

11  :00 

22 

46 

12:00           70                57 

12  :00 

20 

45 

§23] 


GRAPHIC  REPRESENTATION 


33 


13.  What  was  the  discharge  at  7  :  00  p.  m.?    At  10  :  00  a.  m.? 

14.  What  was  the  minimum  rate  of  discharge?     At  what  time? 
16.  What  was  the  water  temperature  when  the  discharge  was  a 

minimum? 

16.  What  was  the  maximum  water  temperature?     At  what  time? 

17.  What  was  the  maximum  air  temperature?     At  what  time? 

18.  What  was  the  greatest  difference  between  the  air  temperature 
and  the  water  temperature? 

19.  At  what  hour  did  this  "maximum"  difference  occur? 

20.  What  was  the  minimum  difference  between  the  air  temperature 
and  the  water  temperature?  ' 

21.  At  what  hour  did  this  "minimum"  difference  occur? 

22.  Can  you  think  of  any  cause  which  would  produce  the  indenture 
in  the  air  temperature  curve  at  1:30  p.  m.,  at  which  time  a  normal 
temperature  curve  should  have  a  maximum? 

23.  Can  you  think  of  any  reason  why  the  discharge  curve  should 
have  a  minimum  when  the  temperature  curves  have  maxima? 


TABLE  V.— DISCHARGE  OF  A  SEEPAGE  DITCH 


Time, 
Aug.  24,  1905 

Discharge  of  ditch, 
sec.-ft. 

Temp,  of  water, 
°F. 

Temp,  of  air, 
0  F. 

8  :00  a.  m. 

3.72 

65 

9:00  a.  m. 
11  :00  a.  m. 
1  :30p.  m. 
2  :  15  p.  m. 

3.70 
3.66 
3.52 
3.49 

70 
79 
83 

74 

84 
85 

3  :  30  p.m. 
5:  30  p.  m. 
6  :00  p.  m. 

3.52 
3.66 
3.73 

84 
78 

90 

88 

8:  00  p.m. 

3.84 

67 

76 

23.  The  Point.  The  location  of  a  point  upon  a  sheet  of  rec- 
tangular coordinate  paper  is  fixed  by  giving  its  distances  from  the 
vertical  and  horizontal  axe^.  The  distance  from  the  vertical  axis 
is  called  the  abscissa,  and  the  distance  from  the  horizontal  axis  is 
called  the  ordinate  of  the  point.  These  two  distances  are  called 
the  coordinates  of  the  point.  The  horizontal  axis  is  commonly 
called  the  X-axis,  or  the  axis  of  x;  and  the  vertical  axis,  the 
Y-axis,  or  the  axis  of  y.  The  abscissa  and  ordinate  are  then 

3 


MATHEMATICS 


[§23 


called,  respectively,  the  x  and  y  coordinates.  When  writing  the 
coordinates  of  a  point,  the  value  of  the  abscissa  is  placed  before 
the  value  of  the  ordinate,  and  separated  from  it  by  a  comma. 
Thus,  "the  point  (3,  4),"  (read,  "three,  four"),  is  a  point  three 
units  to  the  right  of  the  F-axis,  and  four  units  above  the  X-axis. 
The  point  (—  2,  6)  is  two  units  to  the  left  of  the  F-axis  and  six 


1 

r 
1 

(-2,6 

) 

6 

5 

4 

(3,4) 

3 

n 

2 

I 

1 

7     - 

6     - 

5     - 

4     - 

3     - 

2  - 

3  (-2, 

1  O 
1)  - 

1 

l 

2 

3 

4 

5 

6 

7 

. 

2 

(3,-: 

S) 

III 

_ 

3 

IV 

4 

_ 

5 

. 

6 

7 

FIG. 

r' 

17. 

units  above  the  X-axis.  The  point  (—2,  —  1)  is  two  units  to  the 
left  of  the  F-axis  and  one  unit  below  the  X-axis.  The  point 
(3,  —  2)  is  three  units  to  the  right  of  the  F-axis  and  two  units 
below  the  X-axis.  These  four  points  are  shown  on  the  sheet  of 
squared  paper  in  Fig.  17. 

The  two  axes  divide  the  plane  into  four  parts,  called  quadrants. 
They  are  numbered  from  "one"  to  "four"  as  shown  in  Fig.  17. 


§23]  GRAPHIC  REPRESENTATION  35 

Exercises 

1.  Plot  the  following  points:  (a)  (1,  1);  (6)  (2,  3);  (c)  (  -  3,  2); 
(d)    (  -  1,  -  1);    (e)    (  -  2,  -  3);    (J)  (3,  -  4);    (g)    (V2,  1);    (ft) 
C  -  V3,  V5). 

2.  Find  the  distance  of  each   of  the  following  points  from  the 
origin:   (a)  (1,  1);  (6)  (  -  V2,  V2);  (c)  (  -  3,  -  2). 

3.  Find  the  distances  between  the  following   pairs    of   points: 
(a)  (1,  1)   and   (2,  2);    (&)   (  -  2,  3)   and  (  -  5,  -  6);  (c)   (3,  -  5) 
and  (-  1,  0). 

4.  What  is  the  y-coordinate  of  any  point  upon  the  axis  of  a;? 
6.  What  is  the  x-coordinate  of  any  point  upon  the  axis  of  y? 

6.  What  is  the  y-coordinate  of  any  point  upon  a  straight  line 
parallel  to  the  axis  of  x,  three  units  above  it? 

7.  What  is  the  ^-coordinate  of   any  point  upon  a  straight  line 
parallel  to  the  F-axis,  five  units  to  its  left? 

8.  What  relation  exists  between  the  coordinates  of   any  (every) 
point  of  a  line  bisecting  the  angle  between  the  positive  directions 
of   the   two   axes?      Between   the  negative  directions  of  the  two 
axes? 

9.  What  relation  exists  between  the  coordinates  of  any  (every) 
point  of  a  line  bisecting  the  angles  having  the  positive  direction  of  the 
F-axis  and  the  negative  direction  of  the  ^T-axis  as  sides? 

10.  What  relation  would  exist  between  the  x-  and  ^-coordinates  of 
any  (every)  point  of  the  line  in  exercise  9,  if  it  were  raised  five  units 
parallel  to  itself?     If  it  were  lowered  six  units  ? 

11.  What  relation  would  exist  between  the  coordinates  of  any  (every) 
point  of  the  line  in  exercise  9,  if  the  line  were  moved  up  three  units 
parallel  to  itself?     If  it  were  lowered  five  units? 

12.  What  are  the  coordinates  of  the  origin? 

13.  Compute  for  every  10°  temperature   (between  0°  and  200°) 
read    upon    a    Fahrenheit    scale,  the  corresponding    reading    upon 
a  centigrade  scale.     Put  your  results  in  tabular  form,  using  a  sheet  of 
paper,  form  M7. 

14.  Compute   for    every  10°  temperature    (between    —  20°  and 
100°)   read   upon   a   centigrade   scale,  the    corresponding   readings 
upon  the  Fahrenheit  scale.     Put  your  results  in  tabular  form,  using  a 
sheet  of  paper,  form  Ml. 

16.  From  the  table  constructed  in  exercise  13,  convert  the  following 
readings  Fahrenheit  into  readings  centigrade:  (a)  12.6;  (6)  21.7;  (e) 
28.55;  (d)  116.75;  (e)  151.86. 

16.  From  the  table  constructed  in  exercise  14,  convert  the  following 


36  MATHEMATICS  [§24 

readings    centigrade  into  readings    Fahrenheit:    (a)     —  2;  (6)  22.2; 
(c)  61.3;  (d)  62.6;  (e)  75.4. 

17.  If  x  represents  the  number  of  degrees  temperature,  as  read  upon 
a  Fahrenheit  scale,  and  if  y  represents  the  number  of  degrees  of  the  same 
temperature  as  read    upon  a  centigrade    scale,    give   the    equation 
connecting  y  and  x. 

18.  Plot  upon  squared  paper  the  numbers  in  the  tables  constructed 
in  exercises  13  and  14.     Plot  degrees  Fahrenheit  along  the  horizontal 
axis  and  degrees   centigrade   along  the   vertical  axis.     Recompute 
numbers,  if  any,  whose  plotted  points  do  not  fall  upon  a  straight  line 
connecting  the  points  (32,  0)  and  (212,  100). 

19.  Construct  a   table,  for  every  sixty-fourth   of  an  inch,    con- 
verting sixty-fourths  into  decimal  equivalents.     Compute  to  three 
decimal  places. 

20.  Plot  the  corresponding  numbers  found    in  exercise    19  upon 
squared  paper.     Do  all  the  points  plotted  fall  upon  a  straight  line 
passing  through  the  origin? 

21.  Construct  a  table  converting  centimeters,  from  0  to  100,  into 
inches.     One  meter  =  39.37079  inches.     Compute  to  three  decimal 
places.     Plot  the  numbers  of  the  table  upon  squared  paper.     Do  all 
the  points  fall  upon  a  straight  line  passing  through  the  origin? 

24.  The  Double  Scale.  From  the  preceding  exercises  we  have 
seen  that  relations  between  numbers  may  be  represented  by  tables, 
by  equations,  and  by  curves  upon  squared  paper.  A  fourth 
method  of  representing  a  relation  between  numbers  will  be  ex- 
plained by  means  of  an  illustration.  If  x  represents  a  pressure 
read  in  feet  of  water,  and  if  y  represents  the  same  pressure  read 
in  inches  of  mercury  (since  mercury  is  13.55  times  as  heavy  as 
water),  the  relation  between  x  and  y  is 

12x 


13.55 
or 

y  =  0.886  x. 

In  Fig.  18,  the  scale  on  the  under  side  of  the  horizontal  line 
represents  pressure  measured  in  feet  of  water,  the  scale  on  the  upper 
side  represents  pressure  measured  in  inches  of  mercury.  The 
scales  are  so  drawn  that  readings  directly  opposite  the  horizontal 
line  represent  the  same  pressure,  i.e.,  the  length  representing  a  unit 


§25]  GRAPHIC  REPRESENTATION  37 

on  the  upper  scale  is  1.13  times  as  long  as  the  length  represent- 
ing a  unit  on  the  lower  scale.  From  this  double  scale,  one  can  easily 
convert  pressure  readings  in  feet  of  water  into  inches  of  mercury  , 
or  vice  versa. 

Inches  of  Mercury,  y 

1  2  3  4  5 


llllll 

|  l|ill||l.lj   I  11111111.(1111          ,,|l   I  ll|lll,|lll.|l  , 

123456789  10 

Feet  of  Water,  a; 

FIG.  18.  —  Double  scale  showing  relation  between  pressure  measured 
in  inches  of  mercury  and  in  feet  of  water. 

Exercises 

1.  Draw  a  double  scale  connecting  centimeters  and  inches. 

2.  Draw  a  double  scale  connecting  inches  and  tenths  of  a  foot. 

3.  Draw  a  double  scale  connecting  pressure  measurements  in  feet 
of  water  and  pounds  per  square  inch.     A  cubic  foot  of  water  weighs 
62.5  pounds. 

25.  The  Equation  of  a  Curve,  the  Graph  of  an  Equation.  Any 
(every)  point  of  a  line  parallel  to  the  axis  of  x  and  three  units  above 
it  has  an  ordinate  three,  and  no  point  whose  ordinate  is  three  is 
found  off  the  line.  Since  the  equation  y  =  3  is  satisfied  by  the 
coordinates  of  every  point  of  this  line  and  by  the  coordinates  of 
no  other  point,  it  is  said  to  be  the  equation  of  the  line.  The  equa- 
tion of  a  line  parallel  to  the  X-axis  and  ten  units  below  it  is 
y  =  —  10.  The  equation  of  the  X-axis  is  y  =  0.  The  equation 
of  a  line  parallel  to  the  F-axis,  five  units  to  its  right,  is  x  —  5. 

The  equation  of  a  curve  is  an  equation  satisfied  by  the  coordi- 
nates of  every  point  of  the  curve  and  by  the  coordinates  of  no  other 
point. 

The  graph  of  an  equation  is  the  locus  of  a  point  whose  co- 
ordinates satisfy  the  equation.  To  illustrate:  if  we  have  a 
circle,1  radius  5  and  center  at  the  origin  of  coordinates,  its  equation 
is  xz  -+-  y2  =  52.  For,  if  we  draw  from  any  point  of  this  circle  a  line 
to  the  origin  and  a  perpendicular  to  the  X-axis  we  have  formed  a 
right  triangle  whose  legs  are  x  and  y,  and  whose  hypotenuse  is  5; 

1  The  term  "circle"  as  used  in  this  book  means  the  locus  of  a  point  at  a  constant 
distance  from  a  fixed  point. 


38  MATHEMATICS  [§25 

and  if  the  point  is  chosen  off  the  circle,  the  hypotenuse  of  the 
right  triangle  is  no  longer  5.  Thus,  for  points  of  the  circle  we  have 
x2  +  y2  =  25,  and  for  points  off  of  the  circle,  x2  +  y2  **  25.  * 

The  graph  of  the  equation  z2  +  ?/2  =  25  is  a  circle,  radius  5 
and  center  at  the  origin. 

The  equation  of  a  straight  line  passing  through  the  origin  and 
making  an  angle  of  45°  what  the  positive  direction  of  the  axis  of  x  is 
y  =  x.  The  graph  of  the  equation  y  =  x  is  a  straight  line  passing 
through  the  origin,  making  an  angle  of  45°  with  the  positive  direc- 
tion of  the  axis  of  x. 

For  brevity,  instead  of  saying  "the  circle,"  or  "the  straight 
line,"  or  "the  curve" — naming  the  equation — "representing 
the  equation" — we  shall  simply  say  the  circle,  or  the  straight 
line,  or  the  curve — naming  the  curve.  Thus,  the  circle  x2 
+  y2  =  25;  the  straight  line  y  =  x,  and  the  curve  x3  —  3xy  — 
yz  =  1. 

Exercises 

1.  What  is  the  equation  of  the  axis  of  x? 

2.  What  is  the  equation  of  the  axis  of  y? 

3.  What  is  the  equation  of  a  line  parallel  to  the  X-axis,  three 
units  above?     Five  units  above?     One  hundred  units  below?     Five 
units  below?    Fifty  units  below? 

4.  What  is  the  equation  of  a  line  parallel  to  the  axis  of  y,  five 
units  to  the  right?     Ten  units  to  the  right?     Six  units  to  the  left? 
Eighty  units  to  the  left? 

5.  What  is  the  equation  of  a  right  line  passing  through  the  origin 
and  bisecting  the  angle  between  the  positive  directions  of  the  coordi- 
nate axes? 

6.  What  is  the  equation  of  a  straight  line  passing  through  the  origin 
and  bisecting  the  angle  between  the  positive  direction  of  the  axis  of 
y  and  the  negative  direction  of  the  axis  of  a;? 

7.  What  is  the  graph  of  y  =  z?     Of  y  =  -  xl    Of  y  =  x  +  2? 
Of  y  =  x  -  3? 

8.  What    is    the    graph    of    y  =  -  z?     Of  y  =  -  x  +  1?     Of 
y  =  -  x  +  3?    Of  y  =  -  z  -  6? 

9.  What  is  the  equation  of  a  straight  line  making  an  angle  of  45° 
with  the  positive  direction  of  the  axis  of  x  and  cutting  the  F-axis 
five  units  above  the  origin?     Ten  units  above  the  origin?     Six  units 
below  the  origin?     Twenty  units  below  the  origin? 

i  "^  "  is  read  "is  not  equal  to,"  or  "is  different  from." 


§25] 


GRAPHIC  REPRESENTATION 


39 


10.  What  is  the  equation  of  a  straight  line  making  an  angle  of  135° 
with  the  positive  direction  of  the  axis  of  x  and  cutting  the  F-axis  one 
unit  above  the  origin?     Three  units  above  the  origin?     Five  units 
below  the  origin? 

11.  What  is  the  equation  of  the  circle,  center  at  the  origin  and 
radius  1?     Radius  2?     Radius  3? 

12.  Show  that  the  equation  of  a  circle,  radius  5  and  center  at  the 
point  (2,  3),  is  (x  -  2)2  +  (y  -  3)2  =  52,  or  z2  +  y*  -  4x  -  6y  =  12. 

HINT:  Take  any  point,  P,  of  the  circle.  Its  coordinates  are  x  and 
y.  Draw  PA  perpendicular  to  the  axis  of  x.  Call  the  center  of  the* 
circle  C,  and  draw  CB  perpendicular  to  PA.  Draw  PC.  PCB  is  a 
right  triangle,  and 

CB2  +  BP2  =  CP2, 
or 

.    (x  -  2)2  +  (y  -  3)2  =  52, 


x2  -  4x  +  4  +  y2  -  6y  +  9  =  25, 


or 


or 


The  student  will  show  that  the  same  equation  is  obtained  when  the 
point  P  is  to  the  right  of,  and  below  C',  to  the  left  of  the  F-axis ;  below 
the  X-axis.  When  the  point 
P  is  to  the  left  of  the  F-axis 
the  student  must  remember 
that  a;  is  a  negative  num- 
ber; when  below  the  x-axis 
the  y  is  a  negative  number. 

13.  Show  that  the  equa- 
tion  of   a   circle,   radius  5, 
center  at  the  point  (3,  2),  is 
x2  +  y2  -  6x  -  4y  =  12. 

14.  Show  that  the  equa- 
tion  of   a   circle,   radius  5, 
center  at  the  point  (  —  2,  3). 
is  x2  +  y2  +  4x  -  &y  =  12. 

15.  Show  that  the  equation  of  a  circle,  radius  5,  center  at  the  point 
(  -  3,  2),  is  x2  +  y2  +  6z  -  4y  =  12. 

16.  Find  the  equation  of  a  straight  line  passing  through  the  origin 
and  making  an  angle  of  30°  with  the  positive  direction  of  the  X-axis. 
Let  T'OT,  Fig.  19,  be  the  line.     Let  P  =  (x,  y)  be  any  (every}  point  of 


40 


MATHEMATICS 


[§26 


the  line.     Let  PA  be  the  ordinate,  y;  and  let  OA  be  the  abscissa,  x. 
OAP  is  a  30°  triangle,  and  (see  exercise  3,  page  8) 


is  the  equation  of  the  line.  For,  the  coordinates  of  every  point  of  the 
line  satisfy  the  equation  and  the  coordinates  of  any  point  not  of  the 
line  do  not  satisfy  the  equation.  If  the  line  weie  raised  three  units, 

x 
its  equation  would  be  y  =  —7=  +  3;  if  it  were  lowered  ten  units,  its 

V3 


equation  would  be  y  = 


-  10. 


17.  Show  that  the  equation  of  a  right  line  passing  through  the  origin 
and  making  an  angle  of  60°  with  the  positive  direction  of  the  axis  of 
x  is  y  =   \3x.      Show  that  if  this  line  were  raised  six  units  its  equation 
would  be  y  =  V  3x  +  6;  and  if  it  were  lowered  five  units,  its  equation 
would  be  y  =  \3x  —  5. 

18.  Find  the  equation  of  a  straight  line  cutting  the  F-axis  one  unit 
above  the  origin  and  the  X-axis  one  unit  to  the  left  of  the  origin; 
cutting  the  F-axis  one  unit  above  the  origin  and  the  X-axis  one  unit 

to  the  right  of  the  origin; 
cutting  the  F-axis  one  unit 
below  the  origin  and  the 
X-axis  one  unit  to  the  right 
of  the  origin;  cutting  the 
F-axis  one  unit  below  the 
origin  and  the  X-axis  one 
unit  to  the  left  of  the  origin. 

26.  The  Equation  of  the 
Straight  Line.    Let  T'OT, 

Fig.  20,  be  any  straight 
line  passing  through  the  origin.  If,  in  addition  to  stating  that 
the  line  passes  through  the  origin,  its  direction  is  given,  the 
line  is  fixed.  Its  direction  may  be  given  by  the  angle  XOT. 
Let  Q  and  R  be  any  two  points  upon  the  line.  Suppose  a  de- 
scribing point  slide  along  the  line  from  Q  to  R,  its  ^-coordinate 


FIG.  20. 


§26]  GRAPHIC  REPRESENTATION  41 

will  change  by  the  amount  QS  and  its  ^-coordinate  will  change 
by  the  amount  SR.  If  the  line  extends  in  the  first  and  third 
quadrants,  both  QS  and  SR  will  be  positive,  or  both  negative,  de- 
pending upon  whether  R  is  above  and  to  the  right  of  Q,  or  below 
and  to  the  left  of  Q.  If  the  line  extends  in  the  second  and  fourth 
quadrants,  QS  and  SR  will  have  opposite  signs.  If  R  is  above  and 
to  the  left  of  Q,  QS  will  be  negative,  and  SR  positive;  if  R  is  below 
and  to  the  right  of  Q,  QS  will  be  positive  and  SR  negative.  It  is 

SR 

readily  seen,  however,  that  for  any  fixed  line  the  ratio  —  is  con- 

Qb 

stant  (remains  unchanged  as  the  positions  of  Q  and  R  are  changed) . 

SR 

Let  us  abbreviate  this  ratio,  — ,  by  a.    If  the  line  extends  into  the 

QS 

first  and  third  quadrants,  a  is  positive;  if  into  the  second  and 
fourth  quadrants,  a  is  negative.  If  the  line  is  very  nearly  hori- 
zontal, a  is  numerically  small.  If  the  line  is  very  nearly  vertical, 
a  is  numerically  very  large.  For  a  line  making  an  angle  of  30° 
with  the  positive  direction  of  the  axis  of  x,  a  is  1  /  "V3.  For  a 
line  making  an  angle  of  45°  with  the  positive  direction  of  the  axis 
of  x,  a  is  1.  For  a  line  making  an  angle  of  60°  with  the  positive 
direction  of  the  axis  of  x,  a  is  >3.  For  lines  making  angles  of 
120°,  135°,  and  150°  with  the  positive  direction  of  the  axis  of  x, 

a  is,  respectively,  —  Vs,  —  1,  and r_- 

Vs 

If  the  line  were  to  rotate  counter-clockwise  about  the  origin 
from  a  horizontal  position  toward  a  vertical  position,  a'  would  in- 
crease continuously  from  zero  through  all  positive  real  values. 
If  the  line  were  to  rotate  clockwise  from  a  horizontal  position  to- 
ward a  vertical  position,  a  would  decrease  continuously  from  zero 
through  all  negative  real  values.  From  this,  we  see  that  every  line 
passing  through  the  origin  has  an  a,  and  for  every  value  assigned 
to  a  there  corresponds  one  and  only  one  straight  line  through  the 
origin. 

The  direction  of  a  line  is  fixed  by  the  algebraic  value  of  a  as  well 
as  by  the  number  of  degrees  in  the  angle  between  the  line  and  the 
positive  direction  of  the  axis  of  x.  The  numerical  value  of  a  is  a 
measure  of  the  steepness  of  the  line.  It  represents  the  number 


42  MATHEMATICS  [§26 

of  times  faster  a  point  moves  in  a  vertical  direction  than  in  a 
horizontal  direction  as  it  slides  along  the  line,  a  is  called  the 
slope  of  the  line.  The  term  slope  is  analogous  to  the  term 
percent  grade  in  railroad  work.  A  1  percent  grade  means  a 
rise  of  1  foot  for  every  100  feet  of  horizontal  distance,  or 
a  slope  of  1  /1 00.  Again  we  speak  of  the  hydraulic  gradient  of  a 
stream  as  2  feet  per  1000,  or  as  8  feet  per  mile;  these  mean,  re- 
spectively, a  slope  equal  to  2/1000  and  8/5280. 

AP 
Let  P,  Fig.  20,  be  any  point  on  the  line.    Then  -~  .    =  a,  or 

y/x  =  a,ory  =  ax.  This,  then,  is  the  equation  of  a  straight  line 
passing  through  the  origin.1  If  the  line  is  raised  two  units,  its 
equation  becomes  y  =  ax  +  2;  if  r.aised  five  units,  y  =  ax  +  5;  if 
lowered  ten  units,  y  =  ax  —  10.  The  equation  of  a  line  parallel 
to  the  line  y  =  ax,  and  cutting  the  F-axis  b  units  from  the  origin,  is 
y  =  ax  +  b.  If  the  line  cuts  the  F-axis  above  the  origin,  b  is 
positive.  If  the  line  cuts  the  F-axis  below  the  origin,  b  is  negative. 
y  =  ax  +  b  is  called  the  slope  equation  of  the  straight  line,  a  is 
called  the  slope,  and  b  the  y-intercept. 

Exercises 

1.  Give  the  slope  and  y- intercept  for  each  of  the  following  lines: 

(a)  y  =  2x  +  1;  (g)  y  =  -  Bx  +  2;  (ro)  y  =  \x  +  3; 

(b)  y  =  2x  -  3;  (h)  y  =  x  +  1;  (n)  y  =  \x  -  3; 

(c)  y  =  3z  -  6;  (i)  y  =  x  -  1;  (o)  y  =  -  %x; 

(d)  y  =  2x  -  1;  (j)  y  =x+  2;  (p)  y  =  -  fx  -  8; 

(e)  y  =  -  2z  -  2;  (fc)  y  =  -  a;  +  6;  (?)  y  =  -  fz; 

(/)  y  =  -  3x  +  4;          (I)   y=  -x  +  2;  (r)  y  =  -  O.Ola:  -  6. 

2.  In  exercise  1,  what  lines  are  parallel  to  (a)?     To  (c)?     To  (e)? 
To  (/)?     To  (A)?      To  (fc)?     To  (TO)?     To    (p)? 

3.  If  a  point  were  to  move  along  the  lines  of  exercise  1,  so  that  its 
x-coordinate  is  decreased  by  one  unit,  would  the   y-coordinate  be 
increased  or  decreased  and  how  much? 

4.  Find  the  equation  of  a  straight  line  passing  through  the  points 
P,  (3,  2),  and  Q,  (1,  5).     To  solve  the  problem  is  to  find  the  value  of 
a  and  of  6  and  substitute  in  the  equation  y  =  ax  +  b.     As  we  pass 
from  the  point  P  to  the  point  Q,  the  x-coordinate  decreases  two  units 

1  The  student  must  guard  against  getting  the  idea  that  the  slope  of  a  line  is,  in 
general,  y  over  x.     This  is  only  true  when  the  line  passes  through  the  origin. 


§27]  GRAPHIC  REPRESENTATION  43 

and  the  ^-coordinate  increases  three  units.  Then  the  slope,  a,  of 
the  line  is  —  3/2.  Let  R  be  the  point  where  the  line  through  P  and 
Q  cuts  the  F-axis.  Draw  PT  perpendicular  to  the  F-axis,  meeting 
it  at  the  point  T.  Draw  QS  perpendicular  to  PT,  meeting  it  at  S. 
PSQ  and  PTR  are  similar  triangles,  and  we  have 

TP  _  TR 
SP   ~  SQ* 
or 

3  _  TR 

2  ~     3  ' 
or 

TB-\ 

and  b  =  L23,  since  b  =  OT  +  TR.     The  equation  of  the  line  is 

_  3         13  ? 

or 

3z  +  2y  =  13. 

This  work  can  be  checked,  for,  substituting  for  x  and  y  the  co- 
ordinates of  the  points  P  and  Q,  the  equation  is  satisfied  in  each  case. 

6.  Does  the  line  of  exercise  4  pass  through  the  point  (2,  7/2)? 
The  answer  is  yes,  for  substituting  2  for  x  and  7/2  for  y,  in  the  equation 
of  the  line,  the  equation  is  satisfied. 

6.  Find  the  equation  of  the  straight  line  passing  through  the  points 
(a)  (-3,  2)  and  (2,  1);  (6)  (1,  1)  and  (2,    2);  (c)    (-1,    3)    and 
(2,  -  3);  (rf)  (-  3,  -1)  and  (-1,  3);  (e)  (-1,2)  and  (6,  -  1). 

7.  Which  lines  of  exercise  6  pass  through  the  point  (—  2,  6)? 

8.  Find  the  equation  of  a  straight  line  passing  through  the  origin 
and  the  point  (3,  5);  through  the  points  (3,  —  1)  and  (—  3,  —  1); 
through  the  points  (—  3,  5)  and  (—  3,  2). 

27.  The  Graph  of  an  Equation  of  the  First  Degree  between  x 
and  y.  In  §  26  it  was  shown  that  every  equation  -of  the  form 
y  =  ax  represents  a  straight  line  passing  through  the  origin;  and 
further,  when  the  line  was  moved  up  or  down  its  equation  became 
y  =  ax  +  b,  where  6  represents  the  algebraic  distance  the  line  was 
translated  in  the  y  direction.  If  the  line  were  moved  from  a  very 
low  position  to  a  very  high  position,  the  b  would  take  on  all  real 
values  between  its  very  small  initial  value  and  its  very  large  ter- 
minal value.  Thus,  by  moving  the  line  over  a  greater  and  greater 


44  MATHEMATICS  [§27 

Vertical  distance,  b  can  be  made  to  take  any  real  value  whatever. 
Any  equation  then  of  the  form  y  =  ax  +  b,  for  any  values  given 
to  a  and  b,  represents  a  straight  line.  It  is  readily  seen  that  this 
form  of  the  equation  includes  all  lines  excepting  those  parallel 
to  the  F-axis.  The  equation  of  a  line  parallel  to  the  F-axis  is, 
however,  x  =  k,  where  the  constant  k  is  the  number  of  units 
the  line  is  distant  from  the  F-axis.  The  equation  of  any 
straight  line  is  then  either  of  the  form  y  =  ax  +  6  or  of  the 
form  x  =  k.  Since  any  equation  of  the  first  degree  between 
x  and  y,  Ax  +  By  =  C,  may  be  put  in  one  of  these  forms,  it  is 
the  equation  of  a  straight  line,  and  is  called  a  linear  equation. 
To  illustrate:  3x  +  %  +  6  =  0  may  be  written  y  =  —  fx  —  3, 
the  equation  of  a  straight  line  having  the  slope  —  f,  and  a 
y-intercept  of  —  3. 

Exercises 
Find  the  slope  and  intercept  of  each  of  the  following  lines: 

1.  2z  -  3y  =  1.  4.  3x  -  2y  +  6  =  0. 

2.  3x  -  2y  -  6  =  0.  5.  y  -  3x  =  5. 

3.  x  -  y  =  1.  6.  y  +  x  =  0. 

7.  Find  the  point  of  intersection  of  each  line  given  above  with  the 
axis  of  x.     (The  distance  from  the  origin  to  the  point  of  intersection 
of  a  line  with  the  X-axis  is  called  the  re-intercept.) 

HINT:  This  is  done  by  putt'ng  y  equal  to  zero,  and  solving  the  re- 
sulting equation  for  x.  Why  does  this  give  the  rc-intercept? 

8.  Draw  the  straight  lines  of  exercises  1  to  6  upon  one  sheet  of 
squared  paper.     Use  1  inch  (or  2  cm.)  to  represent  the  unit  and  place 
the  origin  near  the  center  of  the  sheet. 

9.  Find  the  coordinates  of  all  points  of  intersection  of  these  lines 
in  so  far  as  they  intersect  upon  the  sheet  of  squared  paper. 

The  coordinates  of  the  points  of  intersection  are  to  be  read  to  two 
decimal  places.  To  make  this  possible  great  care  must  be  exercised 
in  drawing  the  lines.  Tabulate  your  results.  In  the  first  column 
place  the  numbers  of  the  lines,  as  2-5,  meaning  the  line  in  exercise 
2,  and  the  line  in  exercise  5.  In  the  second  column  place  the 
x-coordinate  and  in  the  third  column  place  the  y-  coordinate  of  the 
point  of  intersection  of  the  two  lines.  A  fourth  and  a  fifth  column 
are  to  be  filled  in  with  the  results  obtained  from  the  next  exercise. 

10.  Consider  all  pairs  of  equations  given  in  exercises  1  to  6  as 
simultaneous  equations  and  solve  for  x  and  y.  Record  the  results  in 
the  table  of  the  preceding  exercise.  Compare  the  corresponding  x 


v 


GRAPHIC  REPRESENTATION  45 

values  and  the  corresponding  y  values  found  by  the  two  methods. 
Should  the  corresponding  values  be  the  same?     Why? 

28.  Analytic  Method  of  Finding  a  and  b.  In  exercise  4,  §  26, 
there  was  described  a  geometric  method  for  finding  the  a  and  the  b 
in  the  equation  of  a  straight  line  passing  through  two  points.  An 
analytic  method  of  finding  a  and  b  will  now  be  illustrated.  Let 
(3,  2)  and  (1,  5)  be  the  points  through  which  the  line  passes.  The 
coordinates  of  each  point  when  substituted  for  x  and  y  in  the  gen- 
eral equation  of  the  line,  y  =  ax  +  b,  must  satisfy  that  equation. 
Thus  we  have,  upon  substitution,  2  =  3a  +  b,  and  5  =  a  +  b. 
These  two  equations  are  called  conditional  equations,  for  they  are, 
in  mathematical  language,  statements  of  the  two  and  only  two 
conditions  imposed  upon  the  line.1  The  first  equation  is  the 
equation  of  condition  which  says  that  the  point  (3,  2)  is  upon 
the  line  y  =  ax  +  b;  the  second  equation  is  the  equation  of  condi- 
tion which  says  that  the  point  (1,  5)  is  upon  the  line  y  =  ax  +  b. 
Solving  these  two  conditional  equations  for  a  and  b  we  ob- 
tain fl  =  —  f,  and  b  =  V.  Substituting  these  values  in 
y  =  ax  +  b,  the  equation  of  the  line  through  the  two  given 
points  becomes  y  =  —  lx  -f  V,  or  3z  +  2y  =  13. 

Exercises 

1.  Find,  by  the  analytic  method,  the  equation  of  the  line  through 
each  of  the  following  pairs  of  points: 

(a)  (1,  2)  and  (-  3,  6);  (c)  (0,  0)  and  (-  1,  -  1); 

(b)  (-2,  6)  and  (3,  -  2);          (d)  (0,  0)  and  (-  2,  -  3). 

2.  Plot  upon  squared  paper  the  data  given  in  Table  VI.     Plot 
hydraulic   gradient2   along   the    F-axis   using   1  cm.    (or    1/2   inch) 

1  A  conditional  equation  may  be  denned  as  an  equation  expressing  some  condition 
given  in  the  problem.     To  illustrate:  If  two  numbers  x  and  y  are  to  be  found  such 
that  their  sum  is  9,  and  the  sum  of  their  squares  is  25,  we  have 

x  +  y  =  9 
and 

x2  +  j/2  =  25. 

The  first  equation  expresses  the  condition  imposed  upon  x  and  y,  that  their  sum 
is  9.  The  second  equation  expresses  the  condition  that  the  sum  of  the  squares  of 
the  numbers  is  25. 

2  By  hydraulic  gradient  is  meant  the  drop  of  the  surface  of  water  in  moving  down 
stream  a  given  distance.     Thus  in  the  table  the  hydraulic  gradient  of  10  feet  per 
thousand  feet  means  a  drop  of  the  water  plane  (water  surface)  at  the  rate  of  10  feet 
for  every   1000   feet    measured   along  the    horizontal   in  the  direction  of   greatest 
decline  of  the  plane. 


46 


MATHEMATICS 


[§28 


to  represent  10  feet.  Plot  velocity  along  the  X-axis,  using  2  cm. 
(or  1  inch)  to  represent  10  feet.  From  the  plotted  points  it 
appears  as  if  the  points  would  fall  (or  would  nearly  fall)  along  a 
straight  line  if  there  were  no  errors  of  observation,  or  if  all  the  readings 
were  taken  under  the  same  conditions.  The  best  that  can  be  done  in 
this  case  is  to  assume  a  linear  relation  between  hydraulic  gradient 
and  velocity.  Draw  with  the  aid  of  a  transparent  triangle  a  straight 
line  among  the  points  which  in  your  judgment  is  the  best  line  to  repre- 
sent this  relation.  Does  the  line  pass  through  the  origin?  Compute 
a  by  taking  two  points  at  or  near  the  ends  of  the  line.  The  student 
must  remember  that  a  is  the  ratio  of  the  number  of  units  in  the  change 
in  the  ^-coordinate  to  the  number  of  units  in  the  change  in  the 
x-coordinate. 

3.  Plot  upon  squared  paper  the  data  for  the  8-inch  well  given  in 
Table  VII.  Plot  head1  along  the  F-axis  using  1  cm.  (or  1/2  inch)  to 
represent  1  foot.  Plot  yield  along  the  X-axis,  using  1  cm.  (or  1/2 
inch)  to  represent  10  gallons.  As  in  the  preceding  exercise  draw  a 
straight  line  among  the  points  and  determine  the  a. 

TABLE  VI. — VELOCITY  OF  WATER  THROUGH  SAND 


Hydraulic 
gradient, 
feet  per  1000 

Velocity 
of  water, 
feet  per  24  hours 

Hydraulic 
gradient, 
feet  per  1000 

Velocity 
of  water, 
feet  per  24  hours 

31.9 

16.9 

21.0 

10.1 

20.8 

11.4 

119.1 

58.8 

54.1 
10  0 

24.5 
4  3 

55.8 

22.7 

TABLE  VII.— YIELD  OF  AN  8-INCH  AND  OF  A  14-INCH  WELL 


Size  of 
well, 
inches 

Head, 
feet 

Yield, 
gallons 
per  minute 

Size  of 
well, 
inches 

Head, 
feet 

Yield, 
gallons 
per  minute 

14 

2.02 

31.4 

8 

2.50 

39.5 

14 

4.00 

59.7 

8 

3.85 

49.4 

14 

6.40 

93.2 

8 

6.47 

89.3 

14 

8.68 

121.1 

8 

8.28 

116.3 

14 

11  73 

151.1 

1  By  "head"  in  this  problem  is  meant  the  distance  the  water  was  lowered  in  the 
well  below  its  normal  position.  Thus,  a  head  of  2.5  feet  in  the  table  means  that  when 
39.5  gallons  of  water  per  minute  were  pumped  from  the  well  the  elevation  of  the  sur- 
face of  the  water  in  the  well  was  2.5  feet  below  its  normal  elevation. 


§29]  GRAPHIC  REPRESENTATION  47 

4.  Plot  upon  squared  paper  the  data  for  the  14-inch  well  given  in 
Table  VII.  Draw  a  straight  line  among  the  points  and  find  its  a.1 

29.  Cost  of  Concrete.  Suppose  concrete  mixture  is  made  by 
mixing  cement,  sand,  and  crushed  rock.  The  expression  "  mixture 
(ra,  n,  fe) "  means  the  relative  amounts,  by  volume,  of  the  three 
materials.  Thus,  a  mixture  (1,  2,  3)  means  that  for  each  volume 
of  cement  there  are  two  volumes  of  sand  and  three  volumes  of 
rock. 

Curves  will  now  be  drawn  from  which  the  cost  of  a  cubic  yard 
of  concrete  may  be  determined.  The  method  of  drawing  the 
curves  will  be  illustrated  by  an  example.  The  student  is  to  con- 
struct the  drawing  as  explained.  Suppose  the  percent  voids  of 
rock  is  45  and  that  of  sand  33.  Let  us  take  the  mixture  (1,  2,  3). 
For  each  cubic  yard  of  cement,  2  cubic  yards  of  sand  and  3  cubic 
yards  of  rock  are  used.  In  the  3  cubic  yards  of  rock  there  are 
3-0.45  =  1.35  cubic  yards  of  voids.  Then  1.35  cubic  yards  of  the 
2  cubic  yards  of  sand  will  be  taken  up  by  the  voids  of  the  rock 
leaving  0.65  cubic  yard.  Thus  mixing  3  cubic  yards  of  rock 
and  2  cubic  yards  of  sand  will  give  but  3  +  0.65  =  3.65  cubic 
yards  of  mixture.  In  the  2  cubic  yards  of  sand  there  are 
2-0.33  =  0.66  cubic  yard  of  voids.  Thus  there  is  an  excess  of 
1  —  0.66  =  0.34  cubic  yard  of  cement  above  filling  the  voids  of 
the  sand.  Hence  in  mixing  the  rock,  sand,  and  cement  there  will 
be  3.65  +  0.34  =  3.99  cubic  yards  of  mixture. 

If  Xi  is  the  cost  (say  in. dollars)  of  a  cubic  yard  of  stone,  the 
cost,  yi,  of  the  stone  in  1  cubic  yard  of  concrete  is 

3 

yi  ~  3^9  Xl 
or 

yi  =  0.752  x,  (1) 

Upon  a  sheet  of  squared  paper  draw  a  straight  line  represent- 
ing equation  (1).  Let  10  cm.  represent  $1.  Then  each  centi- 
meter will  represent  10  cents.  To  draw  this  line  join  the  points 
(0,  0)  and  (2,  1.504). 

1  The  equations  found  in  exercises  2,  3,  and  4  very  probably  do  not  represent  the 
true  relations  among  the  measured  numbers,  and  are  at  best  only  approximate 
relations.  Such  equations  are  called  empirical  equations,  or  empirical  laws. 


48  MATHEMATICS  [§29 

If  x2  represents  the  cost  of  1  cubic  yard  of  sand,  the  cost,  y2, 
of  the  sand  in  1  cubic  yard  of  concrete  is 


2/2  = 


3799    2 
or 

?/2  =  0.501  x2  (2) 

Plot  equation  (2)  upon  the  same  sheet  of  paper  upon  which 
equation  (1)  was  plotted,  using  the  same  scales. 

A  barrel  (four  sacks)  of  cement  contains  3|  cubic  feet,  or  5/36 
cubic  yard.  If  xs  represents  the  cost  of  cement  per  barrel,  the 
cost,  y3,  of  cement  in  1  cubic  yard  of  concrete  is 

36 


z  ~  5-3.99     3 
or 

7/3  =  1.805  x3  (3) 

Draw  this  line  upon  the  same  sheet  of  paper  upon  which  equations 
(1)  and  (2)  were  drawn. 

Mark  the  first  line  drawn  "rock;"  the  second  "sand;"  and  the 
third  "cement."  From  these  three  lines  the  cost  of  material  in 
1  cubic  yard  of  concrete  can  be  scaled  off,  if  the  costs  of  stone, 
sand,  and  cement  are  given.  Thus,  suppose  stone  worth  $1.40  per 
cubic  yard,  sand  $1.25  per  cubic  yard  and  cement  $1.10  per 
barrel.  From  the  point  1.4  on  the  X-axis  go  up  to  the  rock  line, 
thence  to  the  left  to  the  F-axis  and  obtain  the  reading  $1.05,  the 
cost  of  the  stone  in  1  cubic  yard  of  concrete.  From  the  point 
1.25  on  the  X-axis  go  up  to  the  sand  line,  thence  to  the  left  to 
the  F-axis  and  obtain  the  reading  $0.63,  the  cost  of  the  sand  in 
1  cubic  yard  of  concrete.  From  the  point  1.1  on  the  X-axis  go 
up  to  the  cement  line,  thence  to  the  left  to  the  F-axis  and  obtain 
$1.99,  the  cost'  of  the  cement  in  1  cubic  yard  of  concrete. 
The  total  cost  of  material  in  1  cubic  yard  of  concrete  is  then 
$1.05  +  $0.63  +  $1.99  =  $3.67. 

The  following  table  gives  the  number  of  cubic  yards  of  rock, 
cubic  yards  of  sand,  and  barrels  of  cement  in  1  cubic  yard  of 
concrete  based  on  45  percent  voids  in  rock  and  33  percent  voids 
in  sand. 


§30] 


GRAPHIC  REPRESENTATION 


49 


Mixture 

Rock, 
cu.  yds  . 

Sand, 
cu.  yds. 

Cement, 
barrels 

1:2  :3 

0.752 

0.501 

1.805 

1:2   :4 

0.881 

0.446 

1.586 

1:2^:5 

0.922 

0.461 

1.327 

1:3   :5 

0.951 

0.475 

1.125 

Exercises 

1.  Verify  the  above  table. 

2.  Construct  curves  from  which  to  scale  off  the  cost  of  material 
used  in  a  1,  3,  5  mixture  of  concrete. 

3.  From  the  drawing  of  exercise  2,  find  the  cost  of  the  material  in 
1  cubic  yard  of  concrete,  if  rock  is  worth  $1.50  a  cubic  yard,  sand 
$1.10  a  cubic  yard  and  cement  $1.30  per  barrel. 

30.  The  Graph  of  y  =  px2. 

Exercises 

1.  Upon  a  sheet  of  paper,  form  M7,  construct  a  table  per  the 
following  instructions.     Head  the  first  vertical  column  x,  the  second 
x2,  the  third  £x,  the  fourth  |x2,  and  the  fifth  ?x2.     In  the  horizontal 
spaces    in  the   first   column    place  the    numbers    0,    0.2,    0.4,   0.6, 
0.8,    1.0,    1.2,   .    .    .,  4.6,  4.8,  5.O.1     In    the  second  column   place 
the  squares  of  these  numbers,  as  0,  0.04,  0.16,  0.36,  0.64,  1.0,   .    .    ., 
21.16,  23.04,  and  25.00.     In   the  third,  fourth   and   fifth   columns 
place,  respectively,  the  halves,  the  thirds  and  the  fourths  of  the  num- 
bers in  the  second  column.     Head  this  table  "Table  of  Squares." 

2.  Plot  upon  squared  paper  a  curve  for  y  =  x2.     Let  the  X-axis 
run  the  shorter  and  the  F-axis  the  longer  dimension  of  the  sheet  of 
paper.     Choose  the  origin  near  the  lower  center  of  the  page.     Plot 
points  for  negative  values  of  x  as  well  as  for  positive  values.     Make 
use   of  your  "Table  of  Squares."     Let  2  cm.  (or   1  inch)  on  both 
axes  represent  one  unit.     Use  a  very  sharp  pointed  pencil  in  plot- 
ting these  points,  and  exercise  great  care  in  locating  them  as  accurately 
as  possible.     Do  not  draw  a  curve  through  the  plotted  points  until 
all  exercises  of  this  list  have  been  completed. 

3.  Plot  a  curve  for  y  =  \x*.     Follow  directions  given  in  exercise  2. 

4.  Plot  a  curve  for  y  =  $x2.     Follow  directions  given  in  exercise  2. 
6.  Plot  a  curve  for  y  =  Jx2.     Follow  directions  given  in  exercise  2. 
6.  Plot  a  curve  for  y  =  ix2.     Follow  directions  given  in  exercise  2. 

1  The  marks  "  .    .    .  "  mean  "and  so  on,"  indicating  all  corresponding    inter- 
mediate numbers. 

4 


50 


MATHEMATICS 


[§30 


7.  Prom  the  coordinates  of  the  points  of  the  curve  y  =  z2  compute 
the  slope  of  lines  passing  through  the  consecutive  plotted  points.     The 
slope  of  the  secant  lines  will  be  approximately  the  slope  of   the 
tangent  lines  drawn  to  the  curve  midway  between  the  points.     Thus, 
if  we  find  a  slope  of  4.2  for  the  secant  line  drawn  through  the  points 
whose  ^-coordinates  are  2.0  and  2.2,  we  shall  have  the  approximate 
slope  of  the  tangent  line  drawn  to  the  curve  at  the  point  whose 
x-coordinate  is  2.1     Tabulate  your  results. 

8.  Same  as  exercise  7  but  use  curve  y  =  £z2. 

9.  Same  as  exercise  7  but  use  curve  y  =  |z2. 

10.  Same  as  exercise  7  but  use  curve  y  =  |x2. 

11.  Same  as  exercise  7  but  use  curve  y  =  £z2. 

12.  Plot  upon  squared  paper  the  results  of  exercise  8.     Plot    x 
along  the  horizontal  axis,  and  slope,  a,  along  the  vertical  axis.     Draw 
a  straight  line  through  the  origin  and  among  the  plotted  points. 
Measure  the  slope  of  this  line;  call  it  A\. 

13.  Same  as  exercise  12  but  use  results  of  exercise   9.     Call  the 
slope  Az. 

14.  Same  as  exercise  12  but  use  results  of  exercise  10.     Call  the 
slope  A*. 

15.  Same  as  exercise  12  but  use  results  of  exercise  11.     Call  the 
slope  At. 

16.  Divide  A 2  by  f;  As  by   |;  and    A^  by  j.     These  quotients 
and  AI  should  all  be  equal  to  2. 

It  can  be  shown  by  the  method  of  the  Calculus  that  the  slope  of  a 
tangent  line  drawn  to  the  curve  y  =  px2  (p  being  a  constant)  is  2px, 
where  x  is  the  abscissa  of  the  point  of  tangency.1 

1  A  formal  proof  is  given 
here.  Let  P,  (x,  y),  Fig.  21, 
be  any  point,  and  Q  any  second 
point  on  the  curve  y  =  px*. 
Let  k,  (PR),  be  the  difference 
of  the  two  z-coordinates,  and 
let  h,  (RQ),  be  the  difference  of 
the  two  ^-coordinates.  The 
coordinates  of  Q  are  then  (x  + 
k,  y  +  h) .  For  the  point  P  we 
have  y  =  px*;  and  for  the  point 
Q  we  have  (y  +  h)  =  p(x  + 
k)t.  Subtracting,  we  have  h 
=  2pkx  +  pk*.  Dividing  by 
k  we  have  h/k  =  2px  +  pk. 
This  shows  that  the  slope, 
(h/k),  of  the  secant  line  drawn 
through  the  points  P  and  Q  is 


§30]  GRAPHIC  REPRESENTATION  51 

This  is  only  saying  that  if  a  point  move  along  the  curve  its  upward 
motion  (downward  motion  if  the  point  is  to  the  left  of  the  y-axis)  is 
at  any  instant  2px  times  its  horizontal  motion,  or  its  y-coordinate  is 
changing  2px  times  as  fast  as  its  x-coordinate.  Here  x  represents  the 
x-coordinate  of  the  position  of  the  moving  point. 

Experiment  shows  that  a  body  falling  freely  from  rest  gives  the  law 
s  =  16.  li2,  where  s  represents  the  distance,  in  feet,  passed  over  in 
the  time  t,  in  seconds.  If  s  be  plotted  along  the  vertical  axis  and  t 
along  the  horizontal  axis  the  slope  of  the  tangent  line  drawn  to  the 
curve  will  be  32.2t.  Distance  s,  then,  is  changing  32.2t  times  as  fast 
as  t,  i.e.,  the  speed  of  a  body  falling  freely  from  rest  is  32.2J,  when  t 
is  the  time,  measured  in  seconds,  elapsed  from  the  instant  the  body 
began  to  move. 

If  v  represents  speed,  in  feet  per  second,  we  have  v  =  32. 2t. 
Plotting  v  along  the  vertical  axis  and  t  along  the  horizontal  axis 
we  obtain  a  straight  line  representing  the  relation  between  speed 
and  time.  The  slope  of  his  line  is  32.2,  i.e.,  v  changes  32.2  times  as 
fast  as  t,  or  the  acceleration  of  gravity  is  32.2  feet  per  second  per 
second. 

17.  Plot  a  curve  for  y  =  x2  —  2x  +  1  =  (x  —  I)2.     How  is  this 
curve  related  to  the  curve  y  =  x2? 

18.  Plot  a  curve  for  y  =  x*  -  2x  +  2  =  (x  -  I)2  +  1.     How  is 
this  curve  related  to  the  one  plotted  in  exercise  17? 

19.  Plot    a    curve    for    3y  =  x*  -  4x  +  13  =  (x  -  2)2  +  9,    or 
y  =  3(2  —  2)2  +  3.     How   is   this   curve  related   to  the  curve  for 
y  =  *x2? 

20.  Without  plotting,  compare  the  following  curves: 
(a)  y  =  x2  and  y  =  —  x2. 

(6)  y  =  |x2  and  y  =  -  £x2. 

(c)  y  =  |x2  and  y  =  —  \xl. 

(d)  y  =  lx2  and  y  =  —  }x2. 

(e)  y  =  (x  -  I)2  and  y  =  -  (x  -  I)2. 


2px  +  pk.  Aa  the  point  Q  slides  along  the  curve  and  approaches  P,  k,  (PR),  grows 
smaller  and  smaller  and  approaches  zero;  the  slope  of  the  secant  line  then  ap- 
proaches 2px.  But,  as  Q  approaches  P  the  secant  line  rotates  about  the  point  P 
and  approaches  the  tangent  line  PT.  Therefore,  the  slope  of  the  tangent  line  is 
the  limit  of  the  slope  of  the  secant  line,  i.e.,  2px. 


52  MATHEMATICS  [§31 

21.  Have  the  curves  considered  in  exercise  20  lines  of  symmetry? 
Give  the  equation  of  the  line  of  symmetry  for  each  curve.1 

22.  Change     each     of     the    following     equations     to    the   form 
y  =  p(x  —  q)z  +  r,  where  p,  q,  and  r  are  constants,  and  find  the  coor- 
dinates of  the  highest  or  lowest  point  on  the  curve: 

(a)  4x2  +  By  =  3x  +  5, 

or  3y  =  -  4x2  +  3x  +  5, 

or  -  |  y  =  x2  —  f  x  —  I , 

or  -  f  y  =  x2  -  f  x  +  V*  -  f  -  &, 

or  -  \y  =  (x  -  f)2  -ff, 

or  y  =   -  f  (x  -  f  )2  +  ||. 

P  =  -  I;  9  =  1;  r  =  If. 

The  coordinates  of  the  highest  point  are  f  and  If. 

(6)  y  =  x*  +  4x  -  6.  (f)  x*+y  =  6. 

(c)  3y  =  x2  +  6x  +  10.  (0)  x2  +  2z  +  y  +  1  =  0. 

(d)  3y  =  4x2  +  6x  +  10.  (h)  2x2  -  5x  -  6y  +  10  =  0. 

(e)  4x2  =  y  -  2x  -  6.  (i)  5x2  +  2x  -  y  =  0. 

31.  The  Parabola.  All  curves  drawn  and  discussed  in  the 
preceding  section  have  equations  reducible  to  the  form 
y  =  p(x  —  q)2  +  r.  They  have  a  line  of  symmetry,  x  =  q,  and 
change  from  an  increasing  to  a  decreasing  curve  or  vice  versa  at 
the  point  (q,  r).2  Each  curve  is  the  curve  y  =  px2  translated 
horizontally  a  distance  q,  and  vertically  a  distance  r.3  Any  curve 
which  is  of  the  shape  of  y  =  px2  is  called  a  parabola.  The  line  of 
symmetry  of  the  curve  is  called  the  axis  of  the  parabola,  and  the 
point  of  intersection  of  the  curve  with  its  axis  is  called  the  vertex.4 

1  A  line  such   that  line  segments   drawn  perpendicular  to  it  and  terminated  by 
the  curve,  are  bisected  by  it  is  called  a  line  of  symmetry  of  the  curve.     Thus  any 
line  passing  through  the  center  of  a  circle  is  a  line  of  symmetry  for  the  circle. 

A  point  such  that  line  segments  drawn  through  it  atid  terminated  by  the  curve, 
are  bisected  by  it,  is  called  a  point  of  symmetry.  Thus,  the  center  of  a  circle  is  a 
point  of  symmetry. 

2  A  curve  is  said  to  be  increasing  when  its  ordinate  increases  as  its  abscissa  in- 
creases; and  decreasing  when  its  ordinate  decreases  as  its  abscissa  increases.     The 
increase  in  each  case  being  understood  to  be  the  algebraic  increase. 

8  Distance  here  means  algebraic  distance.  When  q  is  positive  the  curve  is  moved 
to  the  right,  when  negative  to  the  left.  When  r  is  positive  the  curve  is  moved  up, 
when  negative  down. 

4  It  can  be  shown  that  the  parabola  is  the  locus  of  a  point  moving  in  a  plane  so 
that  its  distance  from  a  fixed  line,  the  directrix,  is  always  equal  to  its  distance  from 
a  fixed  point,  the  focus.  The  parabola  is  also  the  curve  of  intersection  of  a  right  cir- 


§31]  GRAPHIC  REPRESENTATION  53 

Any  equation  of  the  form  y  =  ax1  +  ftx  +  7,  where  a,  /3  and 
7  are  constants,  positive  or  negative,  represents  a  parabola  with 
its  axis  parallel  to  the  Y-axis;  for 


ax2  +  0x  +  y  =  a 


which  is  of  the  form  p(x  —  q)2  +  r,  where  p  =  a,  q  =  -  —  >  and 

2a 


r  =  --  -  ----     The  expresson  ax2  +  fix  +  7  is  the  most  general 

expression  of  the  second  degree  in  x,  or  it  is  a  general  quadratic. 
Thus  we  see  that  the  equation  y  =  ax2  +  0x  +  7  represents  a 
parabola  with  axis  parallel  to  the  Y-axis,  and  by  completing  the 
square  in  the  x  terms  we  are  able  to  locate  its  vertex.  Problems 
illustrating  this  point  have  already  been  given  in  exercise 
22,  §  30. 

We  have  already  seen  that  the  slope  of  the  tangent  line  drawn  to 
the  curve  y  =  px2  is  2px,  i.e.,  2p  times  the  x-coordinate  of  the 
point  of  contact  of  the  tangent  line.  Another  way  of  saying  this  is 
that  the  y  is  changing  2px  times  as  fast  as  the  x.  From  this  and 
what  was  said  in  the  preceding  paragraph  we  see  that  the  slope 
of  the  tangent  line  drawn  to  the  curve  for  y  =  ax2  +  /3x  -\-  y  is 
2p(x  —  q),  or  2p  times  the  distance  of  the  point  of  contact  of 
the  tangent  line  from  the  axis  of  the  parabola.  Conversely,  it 
may  be  shown  that  if  a  number  y  changes  2p(x  —  q)  times  as  fast 
as  the  number  x,  x  and  y  are  connected  by  the  parabolic  law. 

Exercises 

1.  Plot  data  given  in  Table  VIII.  Plot  temperature  along  the 
horizontal  axis  and  weight  along  the  vertical  axis. 

cular  cone  and  a  plane  parallel  to  an  element  of  the  cone.  The  parabola  has  many 
useful  and  interesting  properties,  one  of  which  is:  If  a  parabola  be  rotated  about  its 
axis  forming  a  surface  of  revolution,  the  paraboloid  of  revolution,  rays  of  light  passing 
out  from  its  focus  are  reflected  by  the  surface  into  lines  parallel  to  the  axis  of  the 
paraboloid.  This  property  is  made  use  of  in  the  construction  of  reflectors. 


54 


MATHEMATICS 


[§31 


2.  An  empirical  equation  of  the  second  degree  in  T  for  the  data 
given  in  Table  VIII  is  W  =  0.06T2  -  0.3677  +  0.47.  Here  W  is 
used  in  order  to  distinguish  between  weight  computed  by  formula  and 
observed  weight,  W. 

Copy  Table  VIII  upon  a  sheet  of  paper,  form  M7.  In  a  third 
column  headed  W  place  weights  computed  by  the  empirical  formula. 
In  a  third  column  headed  W  —  W  place  the  differences  between  the 
observed  and  computed  weights.  Plot  W  upon  the  drawing  of 
exercise  1.  Mark  the  points  representing  W  so  that  they  are  readily 
distinguished  from  the  points  representing  W. 

TABLE  VIII.— RELATION  BETWEEN  TEMPERATURE  AND  WEIGHT  OF 
10  LITERS  OF  WATER 

T  =  temperature    in    degrees    centigrade. 

W  =  loss  in  weight,  measured  in  grams,  of  10  liters  of  water,  as  temperature 
differs  from  4°. 


T 

w 

T 

w 

0 

1.3 

16 

10.3 

2 

0.3 

18 

13.8 

4 

0.0 

20 

17.7 

6 

0.3 

22 

22.0 

8 

1.2 

24 

26.8 

10 

2.7 

26 

31.9 

12 

4.8 

28 

37.1 

14 

7.3 

30 

43.3 

3.  Plot  data  given  in  Table  IX.1  Plot  ^  along  the  horizontal 
axis  using  1  cm.  to  represent  one-tenth  unit.  Plot  velocities 
along  the  vertical  axis,  using  1  cm.  to  represent  two  one-hun- 
dred ths  of  1  foot  per  second.  Sketch  a  smooth  curve  among  the 
points.  The  curve  may  not  pass  through  all  the  plotted  points.2 
Begin  to  number  the  vertical  axis  with  2.90.  The  true  origin 
will  be  far  below  the  sheet  of  paper. 

1  The  velocity  at  any  point  of  a  moving  stream  is  determined  by  a  current  meter 
placed  at  that  point. 

2  This  curve  is  called  a  vertical  velocity  curve.     In  practical  work,  however, 
velocities  are  plotted  along  the  horizontal  axis  and  depths  along  the  vertical  axis,  and 
down  from  the  origin.     Your  drawing  gives  the  usual  form  of  plotting  if  it  is  turned 
90°  in  a  clockwise  direction.     Vertical  velocity  curves  are  parabolic  in  shape,  with 
the  axis  of  the  parabola  parallel  to  the  surface  of  the  water. 


§32] 


GRAPHIC  REPRESENTATION 


55 


TABLE  IX.— RELATION  BETWEEN  VELOCITY  AND  DEPTH  AT  A 

POINT  IN  THE  LOWER  MISSISSIPPI 
Depth  at  observed  point  =  d;  whole  depth  =  D. 


d 

D 

Velocity,  feet  per 
second 

d 

D 

Velocity,  feet  per 
second 

0.0 

3.195 

0.5 

3.228 

0.1 

3.230 

0.6 

3.181 

0.2 

3.253 

0.7 

3.127 

0.3 

3.261 

0.8                        3.059 

0.4 

3.252 

0.9 

2.976 

4.  An  empirical  formula  connecting  velocity,  V,  and  depth,  x, 
is  V'  =  -  0.78(x  -  0.296)2  +  3.261.  Compute  V  from  this  formula 
and  tabulate  your  results  following,  in  a  way,  directions  given  in 
exercise  2. 

6.  A  strip  of  tin  L  feet  long  and  40  inches  wide  is  made  into  a  trough 
with  rectangular  cross  section  by  bending  up  an  equal  portion  of  each 
side.  Find  the  width  of  each  portion  bent  up  such  that  the  volume  of 
the  trough  is  a  maximum. 

HINT:  The  volume  is  a  maximum  when  the  area  of  the  cross  sec- 
tion is  a  maximum.  Let  y  be  the  area  of  the  cross  section.  Let  x  be 
the  width  of  the  portion  bent  up.  The  dimensions  of  the  cross  section 
are  then  x  and  40  —  2z.  Hence 

y  =  x(40  -  2x) 

y  =  -  2(z2  -  20x) 

y  =  -  2(x2  -  2Qx  +  100)  +  200 

y  =  -  2(»  -  10)2  +  200 

The  equation  is  that  of  a  parabola  with  its  high  point  at  (10,  200). 
Thus  the  maximum  area,  200,  occurs  when  10  inches  of  tin  are  turned 
up. 

6.  Same  as  exercise  5,  but  with  the  width  60  inches  and  length 
equal  to  10  feet.  Find  the  volume  of  the  trough. 

32.  The  Equilateral  Hyperbola. 

Exercise 

1.  Plot  upon  squared  paper  a  curve  for 

123 
(a)  y  =  -  (6)  y  =  ~  (c)  y  -  — 

XXX 


56  MATHEMATICS  [§32 

Choose  the  origin  near  the  center  of  the  sheet  and  plot  for  negative 
as  well  as  for  positive  values  of  x.  Let  1  cm.  (or  1/2  inch)  rep- 
resent the  unit  along  each  axis.  When  x  is  numerically  small  the 
interval  between  successive  values  assigned  to  it  must  be  small  in 
order  to  obtain  points  near  enough  together  for  sketching  the  curve. 
Compare  curves  (a),  (6),  and  (c). 
These  curves  are  equilateral  hyperbolas.  They  are  special 

cases  of  the  more  general  equilateral  hyperbola,  y  =  —'  where    a 

X 

is  a  constant.  When  one  quantity,  y,  varies  with  another,  x, 
so  that  the  relation  connecting  the  two  is  represented  by  the  equa- 
tion y  =  — ,  y  is  said  to  vary  inversely  with  x. 

tC 

If  points  plotted  upon  squared  paper  arrange  themselves  along 
a  curve  which  seems  to  be  an  equilateral  hyperbola,  y  =  - 

X 

may,  as  a  first  assumption,  be  taken  as  an  empirical  equation 
connecting  the  variable  numbers  plotted.  The  problem  now 
becomes  one  of  selecting  a  value  for  the  constant  a,  and  of  deter- 
mining the  agreement  between  the  values  of  the  numbers  plotted 
and  those  called  for  by  the  equation.  If  the  a  cannot  be  selected 
giving  a  fairly  close  agreement  between  observed  and  calculated 
numbers,  it  means  that  the  law  we  are  dealing  with  in  nature 
is  not  that  of  one  number  varying  inversely  as  the  other. 

An  easy  elementary  method  for  determining  the  a,  as  well  as 
showing  whether  or  not  the  law  is  that  of  the  inverse  power,  will 
be  illustrated  by  means  of  an  example. 

In  Table  X  are  given  the  relative  volumes  of  a  mass  of  air 
subjected  to  pressures  ranging  from  0  to  100  pounds  per  square 
inch  measured  above  atmospheric  pressure.  The  temperature 
of  the  air  is  constant,  60°  Fahrenheit. 

Plot  the  data  upon  a  sheet  of  squared  paper.  Plot  pressure 
along  the  F-axis,  using  2  cm.  (or  1  inch)  to  represent  10  pounds. 
Plot  volume  along  the  X-axis,  using  10  cm.  (or  5  inches)  to  rep- 
resent the  unit  volume.  This  curve  seems  to  be  of  the  general 

shape  of  y  =  —  excepting  that  it  crosses  or  touches  the  .X-axis 

X 

at  unity  instead  of  remaining  above  the  X-axis  and  becoming 
nearer  and  nearer  to  it. 


§32] 


GRAPHIC  REPRESENTATION 


57 


To  find  an  equation  connecting  x  (volume)  and  y  (pressure) 
proceed  as  follows.  Construct  a  third  column  of  numbers  for 
Table  X,  by  taking  reciprocals  of  the  numbers  representing  rela- 
tive volumes  given  in  the  second  column.  Call  these  values  the 
x'  values.  (The  prime  is  put  upon  the  x  to  distinguish  it  from  the 
x  which  represents  the  true  volume  reading.)  Plot  a  curve  be- 


100 


90 


•8 

£  70 


#60 


" 


3  40 


30 


20 


10 


Reciprocal 


I  of  Volume 

A 8 


0.2 


0.4        0.6 
Volume 


0.8 


1.0 


FIG.  22. 


tween  pressure  and  x'.  Let  2  cm.  (or  1  inch)  along  the  Y- 
axis  represent  10  pounds;  let  2  cm.  (or  1  inch)  along  the  hori- 
zontal axis  represent  one  unit  of  x' '.  Such  a  curve,  reduced,  is 
given  in  Fig.  22.  This  curve  is  a  straight  line  with  slope  equal 
to  14.7  and  y-intercept  equal  to  —  14.7.  The  equation  of  the 


58 


MATHEMATICS 


[§32 


line  is  y  =  14. 7o/  —  14.7.  Since  x'  —  — ,  the  relation  connect- 
ing volume  and  pressure  is  represented  by  the  equation 

14.7 

y  =  —  - —  14.7.     It  is  readily  seen  that  this  is  the  equation  of 

u/ 

14.7 

the  curve  y  =  -    -  lowered   14.7  units;    that   the    first    curve 
*c 

plotted  between  pressure  and  volume  is  an  equilateral  hyperbola; 
and  that  the  pressure  readings  given  in  Table  X,  when  increased 
by  14.7,  vary  inversely  as  the  volume. 

TABLE  X.— AIR  COMPRESSION  FROM  ONE  ATMOSPHERE  AT 
SEA-LEVEL 


Pressure, 
Ib.  per  sq.  in. 

Volume 

Pressure, 
Ib.  per  sq.  in. 

Volume 

0 

1,000 

45 

0.246 

1 

0.936 

50 

0.227 

2 

0.880 

55 

0.210 

3 

0.830 

60 

0.196 

4 

0.786 

65 

0.184 

5 

0.746 

70 

0.173 

10 

0.595 

75  ' 

0.163 

15 

0.495 

80 

0.155 

20 

0.423 

85 

0.147 

25 

0.370 

90 

0.140 

30 

0.328 

95 

0.134 

35 

0.295 

100 

0.128 

40 

0.268 

The  pressure  readings  given  in  the  table  were  taken  by  means  of 
a  pressure  gauge  which  registers  the  difference  in  pressure  between 
the  outside  and  inside  of  the  receptacle  containing  the  air.  If  we 
assume  Boyle's  law  for  the  compressibility  of  a  gas  (the  volume 
varies  inversely  as  the  pressure)  the  y-intercept,  14.7,  found 
above,  shows  that  when  the  experiment  was  conducted  the  pres- 
sure gauge  was  registering  14.7  pounds  less  than  the  true  pressure 
exerted  upon  the  air,  or  that  the  atmospheric  pressure  was 
14.7  pounds  per  square  inch. 

If,  in  Fig.  22,  the  plotted  points  had  not  arranged  themselves 
along  a  straight  line  the  method  outlined  in  §  28  could  have  been 


§32] 


GRAPHIC  REPRESENTATION 


59 


used  in  finding  the  slope  and  intercept.  When  the  plotted  points 
arrange  themselves  -along  a  curve  which  is  not  a  straight  line,  it 
may  mean  that  the  original  variables  are  not  connected  by  the 
equilateral  hyperbolic  law. 

Exercise 

Find  the  equation  connecting  I  and  W,  using  the  data  given  in 
Table  XI,1  where  7  represents  the  indicated  horse  power  of  a 
steam  engine,  and  W  the  number  of  pounds  of  steam  consumed 
per  hour. 

TABLE  XI 


I 

W 

/ 

W 

36.8 

460.0 

15.8 

222.8 

21.5 

406.2 

12.6 

182.7 

26.3 

344.5 

8.4 

137.0 

21  0 

279  3 

1  Taken  from  Elementary  Practical  Mathematics  by  John  Perry. 


CHAPTER  II 
LOGARITHMS 

33.  Logarithm  Defined.  The  equation  y  =  xz  —  1  is  solved 
for  y.  If  values  are  assigned  to '  x,  the  corresponding  values 
of  y  may  be  calculated.  The  same  equation  may  be  written 
x  =  ±  \/y  -f-  1.  Here  the  equation  is  solved  for  x,  and  if  values 
are  assigned  to  y,  1;he  corresponding  values  ofz  may  be  calculated. 
The  two  equations  represent  the  same  law  connecting  x  and  y. 

Exercises 

1.  Solve  each  of  the  following  equations  for  x: 

(a)y  =  x;  (i)   y  =  3x2  -  2x  +  1; 

(6)  y  =  x  -  1;  (j)  y  =  xs; 

(c)t/  =  2x+3;  (k)y=x*-l; 

(d)y  =  |x  -6;  (I)   y  =  3z»  -3; 

(e)  y  =  x2;  (m)  y  =  x3  +  3x2  +  3x  +  1; 

(/)  y  =  2x^5  (n)  y  =  X3  -  3x2  +  3x; 

(0)  y  =  x2  -  2x  +  1;                      (o)  y  =  ±  Vx*  -  25. 

(h)  y  =  3x2  +  6x  +  3; 

2.  Solve  each  of  the  following  equations  for  x  and  also  for  y: 

(a)  x2  +  y2  =  25;  (i)    xy  =  2; 

(6)  x2  -0«  =  25;  (j)     zy  +y  =  1; 

(c)  x2  +  2x  +  y2  =  25;  (jfe)  xy2  =  3; 

(d)  x2  -  2x  +  y2  =  25;  (I)    x*y  =  5; 

(e)  x2  +  y2  -  2y  =  25;  (m)  x2y  +  xy2  =  3; 

(/)   x2  +  2x  +  y2  -  2y  =  25;       (n)   x2?/  +  xy  +  xy2  =  0; 
(g)   x2  +  x  +  y2  -  2y  =  16;        (o)  xy3  =  1; 
(A)  xy  =  1;  (p)  x=y  =  -  1. 

All  equations  of  the  preceding  exercise  can  be  solved  for  x  by 
means  of  the  symbols  and  operations  of  elementary  algebra.  The 
equation  y  =  ax,  where  a  is  a  known  positive  constant  different 

60 


§33]  LOGARITHMS  61 

from  zero,  cannot,  however,  be  solved  for  x  by  means  of  these 
elementary  operations.  The  best  we  can  do  is  to  say,  x  =  the 
exponent  of  the  power  to  which  a  must  be  raised  to  produce  y. 
In  this  verbal  sense,  the  equation  is  now  solved  for  x.  The 
words  "exponent  of  the  power  to  which  a  must  be  raised  to  pro- 
duce y"  are  abbreviated  by  "logarithm  of  y  to  the  base  a,"  and 
in  writing  they  are  further  abbreviated  by  "Iog0  y."  Thus,  the 
equation  y  =  ax,  solved  for  x,  gives  x  =  loga  y-  Logo  is  then 
a  new  symbol;  considered  as  an  operator,  operating  upon  y,  it 
gives  the  number,  x,  such  that  when  a  has  x  for  an  exponent, 
the  power  equals  y.  a  is  called  the  base.  The  definition  then 
of  the  logarithm  of  a  number  x  to  the  base  a  is:  "the  exponent 
of  the  power  to  which  a  must  be  raised  to  produce  the  number 
in  question." 

Of  the  two  equivalent  equations,  y  =  a*  and  x  =  logo  y,  the 
first  is  called  the  exponential  form,  and  the  second  the 
logarithmic  form. 

Exercises 

1.  Write  the  following  equations  in  logarithmic  form : 

(a)  100  =  102;  (e)  25  =  52;  (i)  1  =  10°; 

(6)10  =  10';  (/)  9  =  (V3)4;  0')2  =  (V2)2; 

\C)    TC    ==    *w    j  \Q ')  -LD    ^=    ~    *j  \fo)  *5    ^~    \'\'   *J  )    j 

(d)  8  =  23;  (h)  125  =  53;  (1)  8  =  (V2)6; 

(m)  2°  =  1;  (n)  5°  =  1. 

2.  Write  the  following  equations  in  exponential  form: 

(a)  Iog24=  2;  (/)  logio  10  =  1;  (ft)  logs  9  =  2; 

(b)  logio  100  =  2;  (0)  logs  3  =  1;  (I)    logio  1000  =  3; 

(c)  logio  1  =  0;  (h)  Iog5  5  =  1;  (m)  logio  10,000=  4; 

(d)  Iog2  1  =  0;  (i)  logs  25  =  2;  (n)   logs  1  =  0; 

(e)  logs  1=0;  (j)  Iog2  32  =5;  (o)    Iog99  1  =  0. 

3.  Give  the  value  of  each  of  the  following: 

(a)  Iog24;  (h)  log,0  100;  (o)  logio  0.1; 

(6)  logs  9;  (i)   logio  1000;  (p)  logio  0.01; 

(c)  logs  27;  (j)  logio  10,000;  (a)  logio  0.001  ; 

(d)  logt  25;  (ft)  logio  1;  (r)  logio  100,000,000; 
(e)log44;  (I)    logs  1;  («)  logio  0.000000001. 
(/)  log«  6;  (m)  logs  1; 

(o)  logio  10;  (n)  logo  1; 


62  MATHEMATICS  [§34 

4.  Evaluate  the  following: 

(a)  Iog22  +  Iog44  +  logB5; 

(6)  logs  9  +  Iog5  25  +  logs  125  +  logs  36; 

(c)  logio  1000  +  Slogs  36-5  Iog2  16+2  logB  625; 

(d)  Iog2  1  -  logs  1  +  logs  3  -  logio  10  +  logs  27; 

(e)  \  logio  100  -  |  logio  10-|  logio  0.1-2  log,0  0.01; 

(/)  logo  a;  (g)  Iog0l 

34.  Theorem :    The  logarithm  of  the  product  of  two  numbers 
is  the  sum  of  their  separate  logarithms. 
Proof:    Let 

logo  u  =  x  and  Iog0  v  =  y 

Then 

ax  =  u          and  a"  =  v 

By  multiplication 

a*+v  =  uv, 
or 

Iog0  uv  =  x  +  y, 
or 

Iog0  uv  =  logo  u  +  logo  v 

Thus  the  theorem :  The  logarithm  of  the  product  of  two  numbers 
is  the  sum  of  their  separate  logarithms. 

It  is  easily  seen  that  the  logarithm  of  the  product  of  three  or  more 
numbers  is  the  sum  of  their  separate  logarithms. 

If  no  base  is  indicated,  the  base  10  is  understood. 

Exercises 

If  log  2  =  0.3010,  log  3  =  0.4771,  log  5  =  0.6990,  and  log  7  =  0.8451, 
find  the  logarithms  of  the  following:1 

(a)     6  (g}  35  (m)     9 

(6)   15  (h)  40  (n)    27 

(c)  10  (t)     4  (o)    18 

(d)  20  (j)     8  (p)    24 

(e)  45                               (k)  16  (q)    30 
(/)    21                                (i)    32  (r)     12. 

1  The  logarithms  used  in  this  book  are  taken  from  a  four-place  logarithmic  table 
and  are  correct  to  the  nearest  fourth  decimal  place. 


§35]  LOGARITHMS  63 

35.  Theorem :  The  logarithm  of  the  quotient  of  two  numbers  is 
the  logarithm  of  the  dividend  diminished  by  the  logarithm  of  the 
divisor. 

Proof:    Let  logo  u  =  x  and  Iog0  v  =  y. 
Then 

a*  =  u  and  a»  =  v. 
By  division 

_  u 
v 
or 

logo  (-)  =  x  -  y, 
or 


logo  y-j  =   10g0  U  —  logo  V. 

Thus  the  theroem:  The  logarithm  of  a  quotient  is  equal  to  the 
logarithm  of  the  dividend  diminished  by  the  logarithm  of  the  divisor. 

Exercises 

Using  the  logarithms  given  in  preceding  Exercise,  find  the  loga- 
rithms of  the  following:  (a)  f ;  (6)  f- ;  (c)  f ;  (d)  I;  (e)  J;  (/)  |. 

36.  Theorem :  The  logarithm  of  the  power  of  a  number  is  the 
logarithm  of  the  number  multiplied  by  the  index  of  the  power. 

Proof:  Let  logo  u  =  x. 
Then 

u—  a1, 
Then 

or 

logo  un  =  nx, 
or 

Iog0  un  =  n  logo  u. 

Thus  the  theorem :  The  logarithm  of  the  power  of  a  number  is 
equal  to  the  logarithm  of  the  number  multiplied  by  the  index  of 
the  power. 

The  theorem  is  true  for  fractional  exponents  as  well  as  for  integral 
exponents,  and  from  it  then  follows  the  law  that  the  logarithm  of  the 


64  MATHEMATICS  [§37 

root  of  a  number  is  equal  to  the  logarithm  of  the  number  divided  by  the 
index  of  the  root. 

ILLUSTRATION  : 

If  log  2_=  0.3010,  log  4  =  2  log  2  =  2  X  0.3010  =  0.6020, 
and  log  V2  =  J  log  2  =  }  X  0.3010  =  0.1505. 

Exercises 

1.  Using  the  logarithms  of  the  number  given  for  the  exercises) 
§  34,  find  the  logarithms  of  the  following:  (a)  32;  (6)  33;  (c)  52;  (d)27; 
(e)  125;  (f)V5;  (g)  3?;  (h)  7*;  (i)  1251;  (j)  625;  (k) 


2.  If  the  log  2  is  0.3010,  what  is  the  (a)  log  20?     (6)  log  200?     (c) 
log  2000?     (d)  log  20,000?     (e)  log  200,000?     How  do  these  loga- 
rithms differ?     What  part  of  these  logarithms  is  the  same?     Why? 

3.  If  log  3  =  0.4771,  what  is  the  (a)  log  30?     (6)  log  300?     (c) 
log  3000?     (d)  log  30,000?     What  part  of  these  logarithms  is  the  same? 
Why? 

4.  If  log  373  =  2.5717,  what  is  the  log  37.3?    log  3.73?  log  3730? 
log  37,300?     What  part  of  these  logarithms  is  the  same?     Why? 
If  a  number  is  greater  than  unity,  how  does  moving  the  decimal 
point  affect  its  logarithm? 

37.  Characteristic  and  Mantissa.  If  10  is  used  as  base,  the 
following  equations  are  true: 

log   10,000  =  4 

log     1,000  =  3 

log        100  =  2 

log          10  =  1 

log  1  =  0 

log  0.1         =  log  1  -  log         10  =  -  1  =  9  -  10 

log  0.01       =  log  1  -  log       100  =  -  2  =  8  -  10 

log  0.001      =  log  1  -  log    1,000  =  -  3  =  7  -  10 

log  0.0001    =  log  1  -  log  10,000  =  -  4  =  6  -  10 

From  the  above  table,  which  can  be  extended  indefinitely  in 
both  directions,  we  see  that  any  number  greater  than  1000  and 
less  than  10,000  has  a  logarithm  lying  between  3  and  4,  i.e.,  the 
logarithm  is  3  plus  some  (decimal)  fraction.  All  numbers  between 


§37]  LOGARITHMS  65 

1000  and  10,000  have  four  digits  preceding  the  decimal  point,  and 
all  numbers,  excepting  1000,  having  four  digits  preceding  the 
decimal  point  lie  between  1000  and  10,000.  Therefore,  the  loga- 
rithm of  any  number,  excepting  1000,  having  four  digits  preceding 
the  decimal  point  is  3  plus  a  (decimal  )fraction. 

In  a  similar  way  it  may  be  shown  that  the  logarithm  of  any 
number  which  is  not  an  integral  power  of  10,  consists  of  an  integral 
part  plus  a  positive  decimal  fraction.  For  fractional  numbers 
the  integral  part  of  the  logarithm  is  usually  written  in  the  form 
of  a  binomial,  as  9  —  10,  8  —  10,  7  —  10,  etc. 

The  integral  part  of  a  logarithm  is  called  the  characteristic. 
The  positive  decimal  part  is  called  the  mantissa.  The  mantissa 
of  a  positive  integral  power  of  10  or  of  the  reciprocal  of  a  posi- 
tive integral  power  of  10  is  zero. 

If  the  log  373  =  2.5717: 

log  3.73  =  log  373       -  log  10  =  2.5717  -  1  =  1.5717 

log  3.73  =  log  37.3       -  log  10  =  1.5717  -  1  =  0.5717 

log  0.373  =  log  3.73     -  log  10  =  0.5717  -  1  =  9.5717  -  10 

log  0.0373  =  log  0.373   -  log  10  =  9.5717  -  10  -  1  =  8.5717  -  10 

log  0.00373  =  log  0.0373-  log  10  =  8.5717  -  10  -  1  =  7.5717  -  10 

The  logarithms  of  numbers  between  1  and  0  are  negative.  It 
will  be  noticed  from  the  above  illustration  of  the  method  of 
writing  negative  logarithms  that  the  mantissa  is  always  the  same 
for  the  same  sequence  of  digits,  and  is  entirely  independent  of  the 
position  of  the  decimal  point. 

The  characteristic  of  the  logarithm  of  a  number  depends  only 
upon  the  position  of  the  decimal  point  in  the  number,  and  is  entirely 
independent  of  the  sequence  of  the  significant  digits.  A  rule  for 
determining  the  characteristic  of  the  logarithm  of  a  number  is: 
the  characteristic  of  the  logarithm  of  a  number,  using  10  as  a  base, 
equals  the  number  of  places  the  first  significant  figure  is  removed 
from  the  units'  place,  and  is  positive  if  the  first  significant  figure  is 
to  the  left  of  the  units'  place,  and  is  negative  if  it  is  to  the  right  of 
units'  place. 

Thus  the  characteristic  of  the  logarithm  of  37.3  is  +  1;  of 
3.73  is  0;  of  0.373  is  -  1,  or  9  -  10;  of  0.0373  is  -  2,  or  8  -  10, 
etc. 

5 


66  MATHEMATICS  [§37 

Exercises 

1.  Give  the  characteristic  of  the  logarithm  of  each  the  following 
numbers  : 

(a)  375;  (e)  87.36;        (ft)  0.6127;      (k)  17.31; 

(6)  172.6;        (/)   8.276;        (t)  0.08217;     (I)    0.00082; 

(c)  37.62;        (g)  0.561;         (j)  0.00756;    (m)  0.000002. 

(d)  186.2; 

2.  Locate  the  decimal  point  in  the  numbers  for  which  the  follow- 
ing are  logarithms: 

(a)  3.7283;    (d)  1.8725;  (g)  3.1786;  (j)  7.6862  -  10; 

(6)  3.6872;     (e)  0.8271;  (ft)  8.1268  -  10;    (/c)  9.3621  -  10. 

(c)  2.7826;    (/)  9.1267-  10;     (i)  0.3675; 

3.  Do  you  see  any  reason  for  using  10  as  a  base  rather  than  any  other 
number?1 

1  When  10  is  used  as  base  the  system  of  logarithms  is  called  the  common  system. 
The  system  of  logarithms  employed  in  theoretical  investigation  has  for  its  base  the 
sum  of  the  infinite  series 

2    I     1     I       1      I        1        1    . 
T  2    T  2-3  ^  2-3-4 

This  number  is  represented  by  the  letter  e,  and  to  seven  decimal  places  is  2.7182818. 
When  the  base  e  is  used  the  logarithm  is  called  the  natural,  the  hyperbolic,  or 
the  Naperian  logarithm.  The  number  e  enters  into  a  great  many  formulas  rep- 
resenting laws  of  nature.  To  illustrate:  if  the  number  of  times  faster  y  is  changing 
than  x  is  proportional  to  y,  it  can  be  shown  that  the  law  connecting  y  and  x  is 
y  =  cekx  where  candk  are  constants.  This  law  is  sometimes  called  the  compound 
interest  law.  The  rate  of  growth  of"  bacteria  is  proportional  to  the  number  of 
bacteria.  The  formula  giving  the  number  at  any  time  is  then  the  compound 
interest  formula. 

It  is  an  easy  matter  to  find  the  logarithm  of  a  number  to  the  base  e,  if  we  know  its 
logarithm  to  the  base  10.  For,  let  logt  N  =  x,  where  N  is  the  number  whose  natural 
logarithm  is  desired.  Then 

ex  =  N 
i_ 

•,,x 

e  =  N    . 

Taking  the  common  logarithm,  we  obtain: 

log  e  =  ~  log  N, 

solving  for  x 

log  AT 


= 
~ 


log  e  ' 
whence 

i         w       log  ^ 
loge  N  =  -.  -  . 
log  e 

The  log  c  =  0.434294.  Then  Ioge  N  =  2.302585  log  N.  This  formula  converts 
the  common  logarithm  of  the  number  N  into  its  natural  logarithm.  To  find  the 
common  logarithm  of  N  if  its  natural  logarithm  is  given  we  use  the  formula, 
log  AT  =  0.434294  loge  N. 


§38] 


LOGARITHMS 


67 


4.  If   the  log   2  =  0.3010,  log    3   =  0.4771,  log   7  =  0.8451,  and 
log  11  =  1.0414,  find  the  logarithms  of  the  following: 

.  121  .    70  .    84V24 

(a)  -v^--  (o)  T-j-p  '"* 

'   15  144 

007 

I  o\      I    I  I  f\ 

21  1.25 


(c) 


16    /  w'  0.064 

5.  Fill  in  the  blanks  of  the  following  table: 


55 

/500\ i 
I  6/  ' 


Number 

Logarithm 

Base 

Number 

Logarithm 

Base 

100 

2 

32 

V2 

125 

5 

81 

5  0 

81 

4 

16 

2.0 

5 

2 

17 

5.0 

27 

3 

V3 

V2 

5 

10 

3V3 

1.5 

625 

5 

38.  Logarithmic  Tables.  A  table  of  common  logarithms  con- 
sists of  the  mantissas  of  the  logarithms  of  numbers  put  in  tabular 
form.  Since  the  characteristic  of  the  common  logarithm  of  a 
numbed  is  dependent  only  upon  the  position  of  the  decimal  point, 
it  is  not  published  in  the  tables,  but  is  supplied  by  the  computer; 
and  since  the  mantissa  is  independent  of  the  position  of  the  decimal 
point,  no  decimal  points  are  placed  in  the  numbers  in  the  table. 
Thus,  if  one  were  to  find  the  logarithm  of  27.6  from  a  table  of 
common  logarithms,  he  would  enter  the  table  with  276,  take 
out  the  corresponding  mantissa,  which  is  a  decimal  fraction,  and 
prefix  the  characteristic  1. 

The  tables  commonly  used  are  known  as  four-place,  five- 
place,  and  six-place;  meaning  by  this  that  the  mantissas  are  given, 
respectively,  to  four,  five,  and  six  decimal  places.  In  a  four- 
place  table  the  number  interval  is  usually  1  in  the  third  place; 
in  a  five-place  table,  1  in  the  fourth  place;  and  in  a  six-place 


68  MATHEMATICS  [§39 

table,  1  in  the  fifth  place.  If  one  wishes  to  compute  accurately 
to  four,  five,  or  six  places,  he  should  use,  respectively,  four-,  five-, 
or  six-place  tables.  Do  not  use  a  five-  or  six-place  table  if  a 
four-place  table  will  suffice  to  give  accurately  the  required 
number  of  places. 

The  use  of  logarithms  and  logarithmic  tables  is  primarily  to 
shorten  the  work  of  numerical  computation,  and  to  relieve  the 
computer  of  as  much  mental  work  as  possible. 

In  the  explanations  of  the  use  of  logarithmic  tables,  it  is  as- 
sumed that  the  student  is  using  Slichter's  four-place  tables.1 
All  numerical  work,  unless  otherwise  stated,  is  to  be  done  accu- 
rately to  four  figures. 

39.  To  Find  the  Logarithm  of  a  Number  from  the  Tables. 
On  the  first  page  of  the  tables  will  be  found  the  logarithms  (i.e., 
the  mantissas)  of  numbers  from  1  to  1000.  There  are  nine  sets 
of  vertical  columns.  The  first  column  of  each  set  is  headed  No. 
and  contains  numbers  whose  logarithms  are  given  directly  op- 
posite in  the  second  column,  headed  log.  In  the  third  column, 
headed  d,  are  found  the  differences  (called  tabular  differences) 
of  the  consecutive  mantissas  given  in  the  second  column.  From 
the  table  one  finds  the  mantissa  for  the  sequence  of  digits  376  to 
be  0.5752.  Hence 


log  3760 

=  3.5752 

log    376 

=  2.5752 

log      37.6 

=  1.5752 

log        3.76 

=  0.5752 

log        0.376 

=  9.5752  - 

10 

log        0.0376 

=  8.5752  - 

10 

log        0.00376 

=  7.5752  - 

10 

Exercises 
Find  the  logarithm  of  each  of  the  following  numbers : 

1.  (a)  365.  (6)  36.5.  (c)  3.65.  (d)  3650. 

2.  (a)  725.  (6)  72.6.  (c)  7.26.  (d)  7260. 

*" Slichter's  logarithmic  and  trigonometric  tables"  are  printed  both  in  pamphlet 
form  and  upon  four  leaves  of  cardboard  bound  together  along  the  longer  edge.  For 
desk  use  the  cardboard  form  should  be  used. 

If  a  student  does  any  considerable  amount  of  numerical  work  he  cannot  afford 
to  use  poorly  arranged  and  slow  tables,  such  as  are  frequently  found  in  text-books  on 
trigonometry  and  algebra. 


§39]  LOGARITHMS  69 

3.  (a)  816.  (6)  81.6.  (c)  8.16.  (d)  8160. 


4.  (a)  261. 

(b)  26.1. 

(c)  2:61. 

(d)  2610. 

6.  (a)  113. 

(b)  1.13. 

(c)  0.113. 

(d)  0.0113. 

6.  (a)  216. 

(6)  0.216. 

(c)  0.00216. 

(d)  216,000. 

7.  (a)  101. 

(6)  1010. 

(c)  0.0101. 

(d)  0.000101 

8.  (a)  2. 

(6)  0.2. 

(c)  0.02. 

(d)  0.00002. 

9.  (a)  9.99.  (6)  99.9.  (c)  0.999.  (d)  9,990,000. 

10.  (a)  0.0136.       (6)  0.000136.     (c)  136,000.        (d)  136,000,000. 

In  writing  a  decimal  fraction  without  an  integral  part  before 
the  decimal  point,  always  place  a  zero  before  the  decimal  point; 
as,  0.1276;  0.0281;  0.1201;  0.0026.  In  writing  and  reading  the 
mantissas  of  logarithms  do  not  omit  the  last  one  or  more  digits, 
even  though  they  be  zero.  Thus,  write: 

log  399  as  2.6010  not  2.601 
log  414  as  2.6170  not  2.617 
log  455  as  2.6580  not  2.658 

In  reading  the  mantissas  of  the  logarithms  of  numbers,  read  two 
digits  at  a  time,  not  one  at  a  time.  Thus,  read  2.7162,  "two, 
(pause)  seventy-one,  sixty-two;"  not,  "two,  point,  seven,  one, 
six,  two." 

If  the  number  consists  of  more  than  three  digits  the  method  of 
finding  the  mantissa  of  its  logarithm  from  the  table  is  illustrated 
by  the  following  examples.  Find  the  log  376.8.  From  the 
table 

log  376  =  2.5752 

log  377  =  2.5763 

The  difference  between  these  logarithms  is  1 1  (disregarding  the 
decimal  point).  This  difference  is  given  in  the  table  in  the  column 
headed  d.  We  see  that  a  difference  of  1  in  the  number  pro- 
duces a  difference  of  11  in  the  logarithm.  A  difference  of 
0.8  in  the  number  will  then  produce  (approximately)  a  difference 
of  11  X  0.8,  or  8.8  (use  9)  in  the  logarithm.  Then,  since  the 
log  376  =  2.5752,  the  log  376.8  =  2.5752  +  0.0009  =  2.5761. 
The  9  added  to  the  logarithm  of  376  is  called  the  correction  for 
8. 

Find  log  521.7. 

log  521  =  2.7168 


70  MATHEMATICS  [§40 

Correction:  9  X  0.7  6 

log  521.7  =  2.7174 
Find  log  8618. 

log  8610  =  3.9350 

Correction  for  8  =4 

log  8618  =  3.9354 

Exercises 

Find  the  logarithms  of  the  following  numbers : 


1.  (a)  172.6. 
2.  (a)  276.5. 
3.  (a)  365.8. 
4.  (a)  812.6. 
6.  (a)  8.887. 

(6)  17.26. 
(6)  27.65. 
(6)  3.658. 
(6)  0.8126. 
(b)  8887. 

(c)  1.726. 
(c)  2.765. 
(c)  0.3658. 
(c)  0.08126 
(c)  0.008887. 

(d)  17,260. 
(d)  27,650. 
(d)  3,658,000. 
(d)  812,600. 

(d)  0.8887. 

If  there  are  five  digits  in  the  number  the  correction  may  be 
found  by  multiplying  the  tabular  difference  by  the  last  two  digits  of 
the  number.  But,  with  the  exception  of  some  of  the  numbers  oc- 
curring near  the  beginning  of  the  table  where  the  tabular  differ- 
ences are  large,  no  error  will  be  introduced  in  the  logarithm  if  the 
number  is  fist  reduced  to  four  significant  figures.  Thus,  to  four 
decimal  places,  the  log  86326  is  the  same  as  log  86330. 

40.  Tables  of  Proportional  Parts.  In  finding  corrections  the 
tabular  difference  is  multiplied  by  the  digit  in  the  fourth  place 
of  the  number,  or  by  the  last  two  digits  of  a  five-place  number. 
This  multiplication  is  performed  by  means  of  the  tables  of  pro- 
portional parts,  the  tables  headed  p.  p.  Thus,  to  find  the  correction 
for  the  log  37,267,  turn  to  the  p.  p.  table  headed  16,  16  being 
the  tabular  difference.  Opposite  6  we  find  9.6,  the  correction  for 

6  in  the  fourth  place;  opposite  7  we  find  11.2,  the  correction  for 

7  in  the  fourth  place.     But  7  is  in  the  fifth  place,  therefore  the 
correction  for  it  is  1.12.     Add  mentally  9.6  and  1.12,  and  obtain 
for  the  complete  correction  10.72  (use  11). 

In  taking  out  the  mantissa  from  the  tables,  the  addition  of 
corrections  should  be  performed  mentally.  A  good  method  to  follow 
is  illustrated  by  example.  To  find  log  13.78.  First  write  down 
the  characteristic,  1.  Then,  with  the  table  at  your  left,  find 
137  in  the  No.  column  and  mark  the  corresponding  mantissa 
by  placing  the  thumb  nail  above,  or  the  finger  nail  below  it. 


§41] 


LOGARITHMS 


71 


Do  not  read  this  mantissa,  but  read  the  tabular  difference,  32. 
From  the  p.  p.  tables  find  the  correction,  26,  for  8.  Now,  go  back 
to  the  mantissa  marked  by  the  finger  nail,  and  read  it  increased 
by  26,  i.e.,  1393.  Then  place  1393  after  the  characteristic  1 
previously  written  down. 

41.  Arrangement  of  Work.  All  work  should  be  arranged  in  a 
vertical  column  and  done  with  pen  and  ink.  Make  the  digits 
about  1/8  inch  tall  and  space  them  horizontally  about  eight 
to  the  inch.  Study  the  formula  in  which  you  are  substituting 
and  decide  upon  some  arrangement  of  your  work  in  the  vertical 
column  which  will  make  the  additions,  subtractions,  etc.,  of 
logarithms  as  systematic  and  easy  as  possible.  Fill  out  the  vertical 
column  as  far  as  possible  before  turning  to  the  table  of  logarithms. 
This  is  called  blocking  out  the  work.  The  method  of  block- 
ing out  the  work  is  illustrated  by  the  following  example:  Find 
the  area,  in  acres,  of  a  triangular  piece  of  land,  the  sides  a,  b 
and  c  being,  respectively,  127.6,  183.7,  and  201.3  rods.  The 
formula  giving  the  number  of  acres,  A,  is1 


A  = 


Vs(s  —  a)(s  —  b)(s  —  c) 


160 


1  The  proof    of   this   formula,  as   generally   given,    presupposes   a    knowledge   of 
trigonometry.     A  proof  involving  only  algebra  and  geometry  is  given  here. 

C 


FIG.  23. 

Let  ABC,  Fig.  23,  be  the  triangle  whose  sides  are  a,  b,  and  c.  Let  CD  be  drawn 
perpendicular  to  AB.     Then 

CD*  -  a*  -  (c  -  AD)*  (1) 

AD  =  \/b*  -  CD*  (2) 

Substituting  in  (1),  

CD*  =  a*  -  c*  +  2c\/b*  -  CD*  -  b*  +  CD*.  (3) 


and 


72  MATHEMATICS  [§41 

where  s  is  one-half  the  sum  of  the  three  sides.     We  may  block  out 
the  work  as  follows: 

s  =  256.3 
a  =  127.6 
6  =  183.7 
c  =  201.3 
2s  =  512.6 
s  -  a  =  128.7 
s  -  b  =    72.6 
s-c=    55.0 
logs  =      2.4087 
log  (s  -a)  =      2.1096 
log  (s  -6)  =      1.8609 
log  (s  -  c)  =      1.7404 
sum  =      8.1196 
l/2sum=      4.0598 
log  160=      2.2041 
log  A  =      1.8557 
A  =    71.73 

In   performing    the  work    the  expressions    on  the   left  of   the 
equality  signs  were  all  written  down  first.     Then  the  numerical 


<j2   _   C2   _   62    =    _   2c&2   -    CD*.  (4) 

Squaring  (4), 

(a*   -    C2   _   62)2    =    4C2(62   _    CD*), 

or 

4c2&2  -  (a2  -  c*  -  b*)*  =  4c2  CD*. 
Factoring, 

(2cb  +  a2  -  C2  -  62)  (2cfe  -  O2  +  62  +  C2)  =  4C2  (7^)2, 
or 

[a*  -  (c  -  6)2][(c  +  6)2  -  02]  =  4c2  CD2. 
Factoring, 

(a  +  6  -  c)(o  -  6  +  c)(c  +  6  +  a)(c  +  6  -  a)  =  4C2  CD*, 
or 

(2s  -  2c)(2s  -  26)  (2s)  (2s  -  2o)  -  4C2  CD*, 
or 

4«(*  -  o)(s  -  6)(s  -  c)  -  c*  CD*, 
or 

cCD 


-  o)(s  -  6)(s  -  c) 


=  area  of  triangle. 


§41]  LOGARITHMS  73 

values  of  a,  b,  and  c  were  filled  in  and  2s,  s,  s  —  a,  s  —  b,  and 
s  —  c  computed.  Then  the  logarithms  of  s,  s  —  a,  s  —  b,  s  —  c, 
and  160  were  taken  from  the  tables.  Then  the  logarithms  of  s, 
s  —  a,  s  —  b,  s  —  c  were  added,  giving  what  is  called  the  "sum." 
Then  one-half  of  this  was  taken  giving  what  is  called  "  1  /2  sum," 
and  from  it  the  logarithm  of  160  was  subtracted  giving  log  A. 
From  the  tables  of  anti-logarithms1  the  number  71.73  was  taken 
out  corresponding  to  the  mantissa  8557,  the  characteristic  1 
fixing  the  decimal  point  after  the  second  digit. 

Exercises 

Compute  the  areas,  in  acres,  of  the  following  triangles,     a,  b,  and 
c  are  expressed  in  rods.     Record  the  time  spent  upon  each  exercise. 

l.o=    93.6  2.  a  =  216.7 

b  =  101.7  b  =  172.5 

c  =  127.3.  c  =  216.3. 

3.  a  =  103.7  4.  a  =  101.73 

b  =  156.4  b  =  127.27 

c  =  217.7.  c  =  131.38. 

[5.]  Plot  a  logarithmic  curve.  Plot  numbers  along  the  X-axis  and 
logarithms  along  the  F-axis.  Let  x  range  from  0.1  to  10.  Let  2 
cm.  (or  1  inch)  on  the  X-axis  and  10  cm.  (or  5  inches)  on  the 
F-axis  represent  one  unit.  Is  this  curve  concave  on  the  upward  or 
on  the  downward  side?  If  the  correction  of  a  logarithm  falls  on  a  0.5 
shall  we  call  it  another  one  or  drop  the  0.5?  Why?  What  would 
be  a  similar  rule  in  taking  out  the  number  from  its  logarithm? 
Why? 

[6.]  With  the  aid  of  the  T-square,  triangle,  compasses,  and  scale, 
draw  to  scale  a  triangle  with  the  sides  given  in  exercise  1.  Let  1/2 
inch  represent  10  rods.  To  do  this  use  the  scale  marked  20  on  the 
triangular  boxwood  scale.  Drop  a  perpendicular  from  any  vertex 
upon  the  opposite  side.  Measure  the  length  of  this  perpendicular. 
Multiply  its  length  (expressed  as  rods)  by  the  length  of  the  base 
(expressed  as  rods)  and  divide  by  320.  How  could  you  get  a  more 
accurate  determination  of  the  area? 

[7.]  Same  as  exercise  6,  but  use  data  given  in  exercise  2. 

[8.]  Same  as  exercise  6,  but  use  data  given  in  exercise  3. 

1 A  table  of  anti-logarithms  is  given  on  the  last  page  of  the  Slichter  table.    From 
it  the  number  corresponding  to  a  given  mantissa  is  obtained.    If  the  use  of  this 
table  is  not  apparent,  the  student  should  read  the  directions  given  on  the  first 
1  cover. 


74  MATHEMATICS  [§42 

[9.]  Same  as  exercise  6,  but  use  data  given  in  exercise  4. 

[10.]  Upon  a  sheet  of  paper  of  good  quality  make  a  drawing  accord- 
ing to  the  following  directions.  Exercise  great  care  in  making  this 
drawing. 

Draw  a  horizontal  line  10  inches  long.  Divide  it  into  ten  equal 
parts  by  drawing  vertical  lines  1/4  inch  long  on  the  lower  side  only. 
Subdivide  each  of  these  ten  intervals  into  ten  equal  parts  by  drawing 
vertical  lines  1/8  inch  long  on  the  lower  side  of  the  horizontal  line. 
We  now  have  a  unit  of  length  (10  inches)  divided  into  100  equal  parts. 
In  other  words,  the  distance  between  any  two  consecutive  vertical 
lines  represents  one  one-hundredth  of  one  unit.  Mark  the  longer 
vertical  lines  beginning  at  the  left-hand  end  0.0,  0.1,  0.2,  0.3,  .  .  ., 
0.9,  1.0.  Upon  the  upper  side  of  the  horizontal  line  and  at  each  end 
draw  a  vertical  line  1/4  inch  long.  Mark  the  one  at  the  left  1,  and 
the  one  at  the  right  10.  Since  the  logarithm  of  1  is  0,  and  of  10  is 

1,  it  will  be  seen  that  the  lines  on  the  upper  side  of  the  horizontal 
line  representing  the  numbers  1  and  10  stand  opposite  their  logarithms. 
The  logarithm  of  2  is  0.3010.     Find  the  location  of  this  number 
(estimating  the  position  for  the  third  place)  on  the  lower  scale,  and 
on  the  upper  side  draw  a  vertical  line  1/4  inch  long.     Mark  this  line 

2.  In  a  similar  way  draw  lines  to  be  marked  3,  4,  5,  .    .    . ,  8,  and  9, 
standing  opposite  the  points  of  the  lower  scale  representing  their 
logarithms.     In  a  similar  manner  draw  vertical  lines  1/S  inch  long 
representing  every  tenth  of  a  unit  between  1  and  2,  2  and  3,  and  so 
on.     We  now  have  a  double  scale.     The  lower  is  uniform,  and  the 
upper,  called  a  logarithmic  scale,  is  non-uniform.     Make  a  second 
drawing  by  placing  the  uniform  scale  above  and  the  non-uniform 
scale,  the  logarithmic  scale,  below. 

11.  Find  the  values  of  the  following,  using  logarithms : 

(a)  (27.83)3(1.621)5.  (6)     V(72.18)3  Vl.382. 

(c)  (0.6382)2  -=-  1.782.  (d)    Vl.278  -h  0.0183. 

(e)  (0.003782)^(1.7821)-.  (/)    0.0007826  H-  0.007936. 

(g)  [3.782  X  26.78  -^  37.92]l          (h)  [3.728  X  17.83  -f-  (2.675)2]*. 

(i)  Vl.186  (0.02763)2  -r-  (0.001726)3. 

12.  Find  the  amount  of  $10.00  in  ten  years  at  compound  interest 
at  6  percent;  (a)  interest  compounded  annually;   (6)  interest  com- 
pounded semi-annually. 

42.  Further  Suggestions  for  Arrangement  of  Logarithmic 
Work.  If  in  a  single  formula  several  substitutions  are  to  be  made, 
similar  operations  should  be  performed  for  each  substitution  before 


§42] 


LOGARITHMS 


75 


another  is   begun.     This   will   be   illustrated   by   the   following 
exercises : 

1.  Table  XII  indicates,  in  part,  the  amount  of  milk  delivered 
at  the  University  of  Wisconsin  Creamery  from  Feb.  16,  1911, 
to  March  15,  1911,  inclusive. 

TABLE  XII 


Pat. 

No. 

Milk, 
Ib. 

Ave. 
test 

Price, 
cents 

Pat. 

No. 

Milk, 
Ib. 

Ave. 
test 

Price, 
cents 

4 

4075 

3.6 

27 

38 

8228 

3.9 

27 

5 

749 

3.5 

27 

39 

2434 

3.85 

27 

2 

2395 

3.8 

27 

63 

6822 

4.00 

27 

3 

4046 

3.6 

27 

80 

1784 

3.45 

26 

6 

1503 

3.6 

26 

65 

1628 

3.7 

27 

7 

2020 

4.15 

27 

102 

1038 

3.7 

27 

8 

9012 

3.6 

26 

105 

705 

4.15 

27 

9 

5750 

3.4 

26 

111 

2646 

3.9 

27 

10 

1491 

3.8 

26 

86 

97 

4.8 

27 

83 

738 

4.05 

27 

44 

1603 

4.6 

27 

36 

2060 

3.6 

27 

81 

2291 

3.65 

27 

37 

2258 

3.8 

27 

82 

111 

4.1 

27 

From  these  data  the  amount  due  each  patron  is  to  be  computed 
by  multiplying  each  hundredweight  of  milk  delivered,  by  the  aver- 
age test  and  then  by  the  price.  For  the  numerical  work  use  a  sheet 
of  computation  paper,  form  M7.  In  the  first  column,  the  column 
marked  0,  on  the  first,  second,  third,  fourth,  fifth,  and  sixth  lines 
write,  respectively,  "pat.  No.,"  "log  of  Ib.,"  "log  of  test,"  "log  of 
price,"  "log  of  amount,"  and  "amount."  Then  on  the  first  line  of 
the  remaining  columns  put  in  from  left  to  right  the  patron's  num- 
ber, as  4,  5,  2,  3,  etc.  Place  one  number  in  each  vertical  column. 
If  there  are  not  enough  vertical  columns  fasten  two  or  more  sheets 
together.  On  the  next  line  fill  in  the  logarithms  of  the  number  of 
hundredweights  of  milk  delivered  by  the  patrons.  On  the  next 
line  fill  in  the  logarithms  of  the  numbers  representing  the  average 
test  of  the  patron's  milk.  On  the  line  below  this  fill  in  the  loga- 
rithms of  the  price  paid  each  patron  per  pound  of  butter  fat.  On 
the  next  line  fill  in  the  sums  of  the  three  logarithms  above.  Then 
take  out  and  place  on  the  next  line  the  anti-logarithms  of  the 


76 


MATHEMATICS 


[§42 


sums,  and  the  numerical  work  is  complete.  By  this  arrangement 
all  numerical  work  pertaining  to  each  patron  is  placed  in  a  ver- 
tical column  headed  by  his  number.  The  order  of  the  work 
should  be  as  outlined  above,  so  that  all  the  operations  of  any 
one  kind  are  completed  before  those  of  another  kind  are  begun. 
As  an  illustration,  the  work  for  patron  No.  65  will  appear  as 
follows: 


65 


1.2117 


0.5682 


9.4314  -  10 


1.2113 


16.27 


2.  A  solid  of  revolution1  is  10  feet  long  and  has  flat  ends  normal 
to  its  axis  of  revolution.  Its  circumferences  at  various  distances 
from  its  larger  end  were  measured  and  the  results  are  given  in 
Table  XIII.  The  numbers  in  the  first  column  represent  the  dis- 

TABLE  XIII 


Distance,  feet 

Circumference,  feet 

Distance,  feet 

Circumference,  feet 

0 

10.00 

5 

6.28 

1 

9.00 

7 

5.47 

2 

8.15 

8 

5.00 

3 

7.40 

9 

4.68 

4 

6.80 

10 

4.40 

tances,  in  feet,  from  the  larger  end  to  the  points  at  which  the 
measurements  were  taken.  The  measurements  are  given  in  the 
second  column.  Compute  the  area  of  the  corresponding  eleven 
cross  sections,  arranging  the  numerical  work  upon  a  sheet  of 
paper,  form  M7,  as  follows:  On  the  first  horizontal  line  in  the 
first  column  of  the  sheet  write  "distance; "  on  the  second  line  write 
"log  cir.;"  on  the  next  line  write  "2  log  cir.;"  on  the  next  line 
write  "log  area;"  and  on  the  line  below  this  write  "area."  In  the 
columns  headed  I,  II,  III,  etc.,  fill  in  from  left  to  right,  on  the  first 
horizontal  line,  the  distances,  as  0, 1,  2,  .  .  .  ,  9,  and  10.  On  the 

1  A  solid  of  revolution  is  the  volume  bounded  by  a  surface  which  may  be  generated 
by  rotating  a  curve  360°  about  a  straight  line. 


§42]  LOGARITHMS  77 

line  below  this  place  the  logarithms  of  the  circumferences.  Multi- 
ply each  of  these  logarithms  by  2  and  place  the  product  in  the 

next  horizontal  line.  Write  the  logarithm  of  -j-  on  the  lower 
edge  of  a  slip  of  paper.  (The  logarithm  of  j-  is  to  be  computed 

TiTT 

separately,  the  work  placed  in  systematic  form  on  the  lower 
portion  of  the  calculation  sheet.)  Place  the  slip  of  paper  so  that 

the  logarithm  of  -r-  stands  directly  over  the  value  of  2  log  cir.  in 

1 
column  I.     Add  logarithm  of  j-  and  the   value  of  2  log  cir., 

placing  the  sum  on  the  fifth  line.  Slide  the  slip  of  paper  to  the 
second  column  and  add  as  before.  Slide  the  paper  to  the  third 
column  and  add,  then  to  the  next,  and  so  on  for  the  eleven  columns. 
Take  out  the  anti-logarithms  corresponding  to  the  logarithms  on 
the  fifth  line,  and  the  numerical  work  is  complete.  Plot  a  curve 
connecting  area  and  distance.  Plot  distance  along  the  X-axis, 
using  2  cm.  (or  1  inch)  to  represent  1  foot.  Plot  square 
feet  along  the  F-axis,  using  2  cm.  (or  1  inch)  to  represent 
1  square  foot.  Find  the  area  under  this  curve  by  counting  the 
squares.  This  area  will  be  proportional  to  the  volume  of  the 
solid.  Why?  What  is  the  volume  expressed  in  cubic  feet? 

3.  BCB'D,  Fig.  47,  represents  the  intersection  of  a  right  circular 
cylinder  with  a  plane  perpendicular  to  the  elements  of  the  cylinder; 
and  BAB' A'  the  intersection  with  a  plane  inclined  30°  to  the  first 
plane.  The  two  planes  intersect  in  the  line  BE'  cutting  the  axis 
of  the  cylinder  at  the  point  0.  Our  problem  is  to  construct  the 
curve  BPAB'.  Take  any  point  R  upon  the  line  B'OB.  Let 
OR  =  x.  Draw  RQ  and  RP,  respectively  in  the  normal  and  in- 
clined planes,  perpendicular  to  BOB'.  P  and  Q,  the  points  where 
the  two  perpendiculars  pierce  the  cylinder,  are  upon  the  same 
element  of  the  cylinder.  PRQ  is  a  60°  right  triangle  with  the 
angle  R  equal  to  30°.  Represent  RP  by  y',  and  RQ  by  y.  Then, 
from  the  property  of  the  60°  triangle, 

V        2 

y      \/3 
or 

y'  =  1.155y. 


78  MATHEMATICS  [§43 


But,  since  BQCB'  is  a  circle,  y  =  \/az  —  xz,  where  a  is  the  radius 
of  the  cylinder.     Then 


y'  =  1.155  Va2  - 


or  y'  =  1.155A/(a  —  x)(a  +  x). 

By  giving  to  x,  (OR),  a  series  of  values  ranging  from  0  to  the 
length  of  the  radius,  a,  the  corresponding  perpendicular  distances 
RP,  (y'},  may  be  computed  and  the  curve  BPAB'  plotted. 

Let  a  equal  2  feet.  Using  the  above  formula,  compute  y'  for 
the  values  x  =  0,  x  =  0.1,  x  =  0.2,  .  .  .  ,  x  —  1.8,  x  =  1.9, 
and  x  =  2.  Plot  a  curve  between  x  and  y',  i.e.,  the  curve  BPAB'.1 

4.  A  barn  is  provided  with  two  3^-foot  diameter  cupola  venti- 
lators connected  with  the  basement.  The  intake  ventilators  in  the 
concrete  basement  wall  are  7  inches  in  diameter,  made  by  using 
7-inch  stovepipe  and  7-inch  elbows  as  forms  in  the  construction  of 
the  wall.  Neglecting  friction,  and  assuming  no  air  leaks  in  around 
the  doors  and  windows,  develop  a  formula  giving  n,  the  required 
number  of  intakes,  so  that  the  velocity  of  the  air  in  the  intakes 
shall  be  the  same  as  the  velocity  in  the  flues  of  the  cupolas. 
Compute  n  without  using  logarithms. 

43.  To  find  the  Anti  -logarithm  from  the  Table  of  Logarithms. 
The  use  of  the  anti-logarithmic  table  is  for  instructional  purposes 
only.  Since  the  method  of  taking  out,  from  an  anti-logarithmic 
table,  the  number  corresponding  to  a  given  logarithm  is  the  same 
as  that  for  finding  the  logarithm  if  the  number  is  given,  the  anti- 
logarithmic  table  enables  the  student  to  become  familiar  with  the 
one  method  before  he  is  to  learn  how  to  use  a  logarithmic  table 
backward,  so  to  speak,  in  finding  anti-logarithms. 

An  anti-logarithmic  table  has  its  advantages,  but  these  are 
more  than  offset  by  advantages  gained  by  using  the  logarithmic 
table  backward.  Five-  and  six-place  tables  are  rarely,  if  ever, 
provided  with  anti-logarithmic  tables. 

1  The  curve  thus  plotted  if  drawn  full  sized  would  be  a  pattern  (not  allowing  for 
the  lap  in  seam)  for  the  saddle  of  a  4-foot  circular  ventilator,  if  the  rafters  of  the 
barn  make  an  angle  of  60°  with  the  vertical. 

Later  an  easier  method  will  be  given  for  drawing  this  curve  as  well  as  a  method 
for  constructing  a  pattern  for  cutting  the  sheet  iron  for  the  flue  of  the  ventilator,  and 
for  the  conical  roof. 


§43]  LOGARITHMS  79 

The  method  of  taking  out  the  number  from  the  logarithmic 
table  which  corresponds  to  a  given  logarithm  will  be  illustrated  by 
example. 

Find  x  if  log  x  =  2.1786.  By  taking  the  mantissa,  1786,  to 
the  table  of  logarithms,  we  see  that  the  number  x  is  between  150 
and  151  (disregarding  decimal  point).  Thus  we  can  write  down 
at  once  the  first  three  digits,  150,  of  the  number  and  whatever 
correction  is  to  be  made  are  digits  suffixed.  The  given  mantissa 
is  25  greater  than  the  mantissa  for  150,  and  the  tabular  differ- 
ence is  29.  Hence  the  required  number  must  be  25  /29  of  one  unit 
greater  than  150.  Since  25  /29  =  0.86,  the  number  corresponding 
to  the  mantissa  1786  is  15086,  or  x  =  150.86.  The  decimal  point 
is  placed  after  the  third  digit  to  correspond  to  the  characteristic  2. 

The  division  of  25  by  29  may  be  effected  by  using  the  propor- 
tional parts  table.  In  the  proportional  part  table  headed  29 
the  nearest  we  can  get  to  25  on  the  smaller  side  is  23.2,  which  cor- 
responds to  8.  This  means  that  in  dividing  25  by  29  the  first 
digit  in  the  quotient  is  0.8,  with  a  remainder  of  25  —  23.2  =  1.8. 
Again  in  the  p.p.  table  we  have  17.4  opposite  6,  which  means 
that  the  nearest  second  digit  of  the  quotient  of  25  by  29  is  6. 
Thus,  at  once  from  the  proportional  parts  table  we  read  that  25 
divided  by  29  is  0.86, 

From  the  above  illustration  the  student  is  not  to  infer  that 
with  four-place  logarithms  one  can  compute  accurately  to  five 
places.  This  is,  in  general,  not  true.  In  general,  only  four  sig- 
nificant figures  of  the  number  should  be  found.  If  the  number 
lies  near  the  beginning  of  the  table  where  the  tabular  differ- 
ences are  large,  the  fifth  digit  may  have  significance  and  should 
then  be  calculated.  In  this  connection  the  student  should  read 
section  72. 

Exercises 

Find,  from  the  table  of  logarithms,  the  numbers  corresponding  to 
the  following  logarithms.  In  finding  corrections  make  use  of  the 
p.p.  tables. 

1.  2.7864.  2.  3.1286.  3.  1.8152. 

4.  9.6278  -  10.     6.  8.1278  -  10.     6.  6.1785  -  10. 


CHAPTER  III 
THE  CIRCULAR  FUNCTIONS:    THE  TRIANGLE 

44.  Angular  Magnitude.  If  a  straight  line,  as  AB,  Fig.  24, 
rotates  about  a  fixed  point  0  until  it  occupies  some  other  (or  the 
same)  position,  as  A'B',  an  angular  magnitude  is  generated, 
namely,  the  amount  of  rotation.  All  angles  are  or  may  be  imag- 
ined to  have  been  generated  in  this  way.  The  initial  position  of 
the  line  is  called  the  initial  side,  and  the  terminal  position  the 

terminal  side  of  the  angle. 
The  point  of  intersection  of 
the  terminal  and  the  initial 
sides  is  called  the  vertex  of 
the  angle.  This  concept  of 
angular  magnitude  is  more 
general  than  that  of  elemen- 
— B  tary  geometry  in  that  the 
angular  magnitude  here  de- 
fined is  not  only  unlimited  in 
YIG.  24.  size,  but  has  a  sense;  i.e.,  the 

angular  magnitude  represents 

either  counter-clockwise  or  clockwise  rotation.  If  the  rotation 
is  counter-clockwise  the  angle  or  the  rotation  is  called  positive; 
if  clockwise,  negative. 

If  we  fix  our  attention  upon  a  particular  spoke  of  a  rotating  fly 
wheel  and  note  its  position  at  the  beginning  and  at  the  end  of  a 
definite  interval  of  time,  we  say  that  this  spoke  has  rotated  through 
an  angle,  the  magnitude  and  sense  of  which  is  the  amount  and 
direction  of  the  rotation  noted.  If  we  had  fixed  our  attention  upon 
a  point  of  the  wheel  and  had  imagined  it  connected  to  the  center  of 
rotation  by  means  of  a  straight  line,  this  imaginary  line  would  have 
rotated  through  the  same  angle  as  the  spoke.  We  say  that  the 
point  has  rotated  through  an  angle.  Since  every  point  of  the 

80 


§44]  THE  CIRCULAR  .FUNCTIONS  81 

wheel  rotates  through  the  same  angle  in  any  given  interval  of 
time,  we  say  the  wheel  has  rotated  through  an  angle. 

The  following  is  another  illustration  of  a  body  rotating  through 
an  angle.  Suppose  the  telescope  of  a  transit  is  first  pointed  at  one 
object  and  then  at  a  second;  we  say  that  the  vertical  planes 
through  the  two  lines  of  sight  form  an  angle  with  each  other.  This 
same  thought  is  expressed  by  saying  that  the  line  of  sight  of  the 
transit  has  been  rotated  through  an  angle.  The  magnitude  and 
sense  of  this  angle  is  the  amount  and  direction  of  rotation. 

The  practical  unit  for  measuring  angular  magnitude  is  the  one 
three  hundred  and  sixtieth  part  of  one  complete  revolution.  This 
unit  is  called  the  degree.  The  degree  is  divided  into  sixty  equal 
parts  called  minutes.  The  minute  is  divided  into  sixty  equal  parts 
called  seconds. 

Since  the  magnitude  and  sense  of  an  angle  is  unchanged  when  it 
is  placed  in  different  positions  of  its  plane,  we  may  move  any  given 
angle  until  its  vertex  coincides  with  the  origin  of  the  system  of 
rectangular  coordinates,  and  until  its  initial  side  coincides  with 
the  positive  part  of  the  X-axis. 

If  the  angle  is  positive  and  less  than  90°  its  terminal  side  will 
fall  within  the  first  quadrant,  and  the  angle  is  said  to  be  in  the 
first  quadrant.  If  the  angle  is  positive  and  greater  than  90°,  but 
less  than  180°,  its  terminal  side  will  fall  within  the  second  quadrant 
and  the  angle  is  said  to  be  in  the  second  quadrant.  If  the  angle  is 
positive  and  greater  than  180°,  but  less  than  270°,  its  terminal  side 
will  fall  within  the  third  quadrant  and  the  angle  is  said  to  be  in  the 
third  quadrant.  Similarly,  a  positive  angle  greater  than  270°  but 
less  than  360°  is  in  the  fourth  quadrant;  if  greater  than  360°  but 
less  than  450°,  in  the  first  quadrant;  and  so  on.  If  the  angle  is 
negative  and  numerically  less  than  90°,  it  is  hi  the  fourth  quadrant ; 
if  numerically  greater  than  90°,  but  less  than  180°,  in  the  third  quad- 
rant; and  so  on.  In  general,  an  angle  is  said  to  be  in  that  quad- 
rant within  which  its  terminal  side  would  fall,  if  its  vertex  were 
placed  at  the  origin,  and  if  its  initial  side  were  placed  on  the  posi- 
tive part  of  the  axis  of  x. 

Thus,  in  Fig.  25,  a,  6,  and  c,  represent  angles  in  the  first  quad- 
rant; d,  e,  and  /,  angles  in  the  second  quadrant;  g,  h,  and  i, 
angles  in  the  fourth  quadrant;  and  j,  k,  and  I,  angles  in  the  third 


82 


MATHEMATICS 


[§44 


quadrant.     In  each  case  the  arrow  indicates  the  sense  and  mag- 
nitude of  the  rotation,    a,  b,  d,  g,  h,  i,  and  k  are  positive  angles. 


FIG.  25. 

Exercises 

1.  In  what  quadrants  are  the  following  angles? 

(a)  16°;      (d)  217°;    (g-)  87°;      (j)  -  76°;      (m)  -  276°; 
(6)  291°;     (e)  796°;     (h)  102°;     (ft)  -  376°;     (n)  -  5°. 
(c)  368°;     (/)  560°;     (i)  212°;      (1)  -  175°; 

2.  With  the  aid  of  the  77-square  and  the  60°  and  the  45°  triangles, 


THE  CIRCULAR  FUNCTIONS  83 

draw  lines  through  the  origin,  making  positive  angles  of  15°,  30° 
45°,  60°,  75°,  90°,  105°,  120°,  135°,  150°,  165°,  and  180°  with  the  positive 
direction  of  the  axis  of  x. 

3.  In  a  60°  triangle  what  is  the  ratio  (a)  of  the  length  of  the 
hypotenuse  to  the  length  of  the  shorter  leg?     (6)  of  the  length  of  the 
hypotenuse  to  the  length  of  the  longer  leg?     (c)  of  the  length  of  the 
shorter  leg  to  the  length  of  the  longer  leg?  (d)  of  the  length  of  the 
longer  leg  to  the  length  of  the  hypotenuse? 

4.  In  a  45°  triangle,  what  is  the  ratio  of  the  length  of  the  hypo- 
tenuse to  the  length  of  a  leg? 

5.  Draw  an  equilateral  triangle  6  inches  on  a  side.     From  each 
vertex  draw  a  line  to  the  middle  of  the  opposite  side.     Mark  the 
numerical  size  of  all  the  angles  formed  by  these  lines. 

6.  Draw  a  right  triangle,  one  acute  angle  of  which  is  60°,  and  the 
hypotenuse  of  which  is  6  inches  in  length.     What  is  the  length  of 
the  shorter  leg  of  this  triangle?     What  is  the  length  of  the  longer  leg? 
What  is  the  ratio  of  the  length  of  the  hypotenuse  to  the  length  of  the 
shorter  leg?     to  the  length  of  the  longer  leg?     What  is  the  ratio  of 
the  length  of  the  longer  leg  to  the  length  of  the  shorter  leg? 

8.  Same  as  exercise  7  but  draw  the  hypotenuse  10  inches  long. 

9.  What  answers  of  exercises  7  and  8  are  the  same?     Will  these 
answers  remain  the  same  for  any  length  of  hypotenuse?     Why? 

10.  Draw  a  right  triangle  with  hypotenuse  equal  to  6  inches  in 
length,  and  an  acute  angle  equal  to  45°.     What  is  the  length  of  the 
legs?     What  is  the  ratio  of  the  length  of  the  hypotenuse  to  the  length 
of  a  leg?     What  is  the  ratio  of  the  length  of  one  leg  to  the  length  of 
the  other  leg? 

11.  Same  as  exercise  10  but  make  the  length  of  the  hypotenuse 
10  inches. 

12.  What  answers  in  exercises  10  and  11  are  the  same?     What 
answers  will  remain  the  same  for  any  length  of  the  hypotenuse? 
Why? 

13.  Draw  a  right  triangle,  one  acute  angle  of  which  is  15°,  and  the 
hypotenuse  6  inches  in  length.     What  is  the  length  of  the  shorter 
leg?     of  the  longer  leg?     What  is  the  ratio  of  the  length  of  the  hypo- 
tenuse to  the  length  of  the  shorter  leg?     of  the  length  of  the  hypo- 
tenuse to  the  length  of  the  longer  leg?  of  the  length  of  the  longer 
leg  to  the  length  of  the  shorter  leg? 

14.  Same  as  exercise  13,  but  make  the  hypotenuse  10  inches  long. 
16.  What  answers  are  the  same   in   exercises  13  and  14?     What 

answers  will  remain  the  same  for  any  length  of   the  hypotenuse? 
Why? 


84  MATHEMATICS  [§45 

45.  Trigonometric  Functions  Denned.  Construct  any  angle  in 
the  first  quadrant.  (This  angle  should  be  general,  and  not  one 
previously  constructed  in  the  preceding  exercises.)  Choose  any 
point,  P,  upon  the  terminal  side,  and  from  it  drop  a  perpendicular 
upon  the  initial  side;  i.e.,  upon  the  X-axis.  We  now  have  a  right 
triangle  with  the  vertex  of  an  acute  angle  (the  original  angle  con- 
structed) at  the  origin.  Th's  triangle  is  called  the  triangle  of 
reference  for  the  constructed  angle.  The  three  sides,  a,  b,  and  c, 
of  this  triangle  may  give  rise  to  six  ratios,  viz. : 

a     b     a    b     c  c 

-'  ~>  -'  -»  -'   and   -• 
c     c     b    a    b  a 

From  a  property  of  similar  triangles  it  is  seen  that  these  ratios  are 
entirely  independent  of  the  size  of  the  triangle;  i.e.,  independent  of 
the  position  of  the  point  P  upon  the  terminal  side,  provided,  how- 
ever, that  the  angle  at  the  origin  does  not  change.  If,  however, 
the  angle  is  made  to  increase  or  decrease  by  any  amount,  no 
matter  how  small,  each  of  the  six  ratios  will  change  in  value.1 
Since  these  ratios  depend  for  their  values  upon  the  size  of  the  angle 
and  are  entirely  independent  of  the  size  of  the  triangle  of  reference, 
they  are  functions2  of  the  angle,  and  are  called  the  trigonometric, 
or  circular,  functions. 

In  the  triangle  of  reference  the  side  opposite  the  given  angle 
(the  angle  whose  vertex  is  at  the  origin)  is  called  the  opposite 
(opp.)  side.  The  side  adjacent  the  angle  is  called  the  adjacent 
(adj.)  side. 

The  six  trigonometric  ratios  are  defined  as  follows,  a  is  the 
angle  in  question: 

opp. 

The  ratio  .  —  is  called  sine  a;  abbreviated  by  sin  a. 
hyp. 

1  The  increase  or  decrease  in  the  size  of  the  angle  should  not  be  so  large  that  the 
angle  will  not  remain  in  the  first  quadrant.     Later  on  it  will  be  seen  that  the  above 
statement  may  not  be  true  if  the  terminal  side  of  the  angle  were  to  move  to  another 
quadrant. 

2  One  quantity,  y,  is  said  to  be  a  function  of  another  quantity  x,  if,  when  x  is  known, 
y  is  determined.     Thus,  the  area  of  a  circle  is  a  function  of  its  radius;  for  if  the  radius 
is  known,  the  area  is  determined.     In  this  case,  since  we  know  the  formula  connecting 
the  radius  and  the  area  of  a  circle,  we  can  actually  compute  the  area.     If   an  angle 
in  the  first  quadrant  is  known,  the  six  ratios  among  the  lengths  of  the   three  sides 
of  the  triangle  of  reference  are  determined. 


§45]  THE  CIRCULAR  FUNCTIONS  85 

The  ratio  ,     '  is  called  cosine  a;  abbreviated  by  cos  a. 
hyp. 

opp. 

The  ratio      ..  is  called  tangent  a;  abbreviated  by  tan  a. 
adj. 

The  ratio         ,  or  the  reciprocal  of  tan  a,  is  called  cotangent 
opp. 

a;  abbreviated  by  cot  a. 

hi/D 

The  ratio  ~TT,  or  the  reciprocal  of  cos  a,  is  called  secant  a; 
duj . 

abbreviated  by  sec  a. 

hwo . 

The  ratio ',  or  the  reciprocal  of  sin  a.  is  called  cosecant  a; 

opp. 

abbreviated  by  esc  a. 

It  is  to  be  remembered  that  in  the  order  given  above,  the  first 
and  last,  the  second  and  fifth,  and  the  third  and  fourth  ratios  are 
reciprocals. 

The  side  opposite  and  the  side  adjacent  are,  respectively,  the 
^-coordinate  and  the  z-coordinate  of  P,  the  point  of  the  terminal 
side  from  which  the  perpendicular  was  let  fall  upon  the  X-axis. 
Since  the  point  P  is  in  the  first  quadrant,  so  that  its  x-  and 
^-coordinates  are  both  positive,  the  side  opposite  and  the  side 
adjacent  of  the  triangle  of  reference  are  considered  positive.  The 
length  of  the  hypotenuse  is  the  distance  of  the  point  P  from  the 
origin,  and  is  always  considered  positive. 

If  the  angle  is  in  the  second  quadrant,  a  triangle  of  reference  is 
formed  as  above  by  dropping  a  perpendicular  from  any  point  P 
of  the  terminal  side  upon  the  X-axis  (the  initial  side  produced 
through  the  vertex) .  The  six  ratios  of  the  lengths  of  the  sides  of 
this  triangle  are  called  the  trigonometric  functions  of  the  angle 
and  are  named  the  same  as  if  the  angle  were  in  the  first  quadrant. 
In  this  case,  however,  the  side  adjacent,  the  z-coordinate  of  the 
point  P,  is  negative;  hence  the  cosine,  the  tangent,  the  cotangent, 
and  the  secant  are  negative. 

In  a  similar  manner,  if  the  angle  is  in  the  third  quadrant,  we  see 
that  both  the  adjacent  and  opposite  sides  of  the  triangle  of  refer- 
ence are  negative,  and  the  sine,  the  cosine,  the  secant,  and  the 


86  MATHEMATICS  [§45 

cosecant  are  negative.     If  the  angle  is  in  the  fourth  quadrant,  the 
sine,  the  tangent,  the  cotangent,  and  the  cosecant  are  negative. 

Exercises 

1.  In  what  quadrant  or  quadrants  is  the  angle  if: 

(a)  The  sine  negative?  (6)  The  tangent  positive? 

(c)   The  secant  negative?  (d)  The  cosine  positive? 

(e)   The  cosecant  positive? 

(/)   The  sine  negative  and  the  tangent  positive? 

(g)  The  secant  positive?  (h)  The  sine  positive? 

(i)    The  tangent  positive  and  the  cosine  negative? 

(j)    The  cosine  negative  and  the  sine  positive? 

(fc)  .The  cosine  negative? 

(I)    The  sine  negative  and  the  cosecant  negative? 

(m)  The  cosine  negative  and  the  cosecant  negative? 

(ri)  The  cosine  positive  and  the  tangent  negative? 

(0)  The  secant  negative  and  the  cotangent  positive?  . 
(p)  The  cotangent  positive  and  the  tangent  positive? 
(q)   The  sine  negative  and  the  tangent  positive? 

(r)   The  cotangent  positive  and  the  cosecant  negative? 
(s)   The  secant  positive  and  the  cosecant  negative? 
(t)    The  cosecant  positive  and  the  tangent  positive. 
(u)  The  cosine  negative  and  the  tangent  positive? 
(v)   The  cotangent  negative  and  the  secant  positive? 
(w)  The  sine  negative  and  the  secant  positive? 
(x)  The  sine  positive  and  the  cosecant  negative? 
(y)  The  tangent  negative  and  the  cotangent  positive? 
(z)   The  cosine  negative  and  the  secant  positive? 

2.  Give  the  values   of  the  six  trigonometric    functions    of    the 
following  angles.     Tabulate  your  results.     Do    not   reduce   radicals 
to  approximate  decimal  fractions,     (a)  30°;  (6)  45°;  (c)  60°;  (d)  120°; 
(e)  135°;  (/)  150°;  (g)  210°;  (h)  225°;  (t)  240°;  (j)  300°;  (k)  315°; 

(1)  330°. 

3.  At  117  feet  from  the  base  of  a   flagstaff  standing    on   level 
ground,  the  angle  of  elevation1  of  the  top  of  the  staff  is  45°;  what  is 

1  If  a  point  A  is  at  a  higher  elevation  than  a  point  B,  the  angle  of  elevation  of  the 
point  A  at  B,  is  the  angle  between  the  horizontal  and  a  line  drawn  from  B  to  A, 
If  a  point  A  is  at  a  lower  elevation  than  a  point  B,  the  angle  of  depression  of  the 
point  A  at  B,  is  the  angle  between  the  horizontal  and  a  line  drawn  from  B  to  A. 


§46]  THE  CIRCULAR  FUNCTIONS  87 

the  height  of  the  staff?  What  is  the  height  if  the  angle  of  eleva- 
tion is  60°?  If  30°? 

4.  In  exercise  15,  page  11  how  high  above  the  plates  is  the 
point  of  meeting  of  the  upper  edges  of  the  upper  and  the  lower 
rafters?  How  far  from  the  outside  of  the  building  will  a  plumb-line 
dropped  from  this  point  strike  the  floor?  How  high  above  the 
plates  is  the  ridge  of  the  barn? 

6.  The  diagonal  of  a  rectangular  plot  of  ground  is  82  rods.  One 
side  is  41  rods.  What  is  the  length  of  the  other  side? 

6.  The  sides  of  a  rectangular  plot  of  ground  are  100  and  173.2 
rods.     What  is  the  approximate  length  of  the  diagonal? 

7.  The  sine   a  =  f ;   the   cosine   is   positive.     Find  the  values 
of  the  remaining  five  functions. 

HINT  :  Since  the  sine  and  cosine  are  both  positive  the  angle  is  in 
the  first  quadrant,  and  all  functions  are  positive.  A  triangle  whose 
hypotenuse  is  five  units  in  length  and  whose  opposite  side  is  three 
units  in  length  (hence  the  adjacent  side  four  units  in  length)  is  a 
triangle  of  reference  for  a.  Hence  we  have,  cos  a  =  -5-,  tan  a  =  f , 
cot  a  =  %,  sec  a  =  -f ,  and  esc  a  =  f . 

8.  Find  the  values  of  the  trigonometric  functions,  construct  to 
scale  a  triangle  of  reference,  and  measure  the  angle  with  a  protractor, 
for  each  of  the  following: 

(a)  sin  a  =  f ,  and  the  cosine  is  negative. 
(6)  cos  a  =  -f ,  and  the  tangent  is  negative. 

(c)  tan  a  =  %,  and  the  sine  is  negative. 

(d)  sin  a  =  —  ^  and  the  tangent  is  negative. 

(e)  sec  a.  =  2,  and  the  tangent  is  negative. 
(/)   sin  a  =  |,  and  the  cosine  is  negative. 
(00  cos  «=•§•,  and  the  sine  is  negative. 
(h)  sec  a  =  -f,  and  the  sine  is  negative. 

9.  At  two  points,  A  and  B,  100  feet  apart  and  in  line  with  the  base 
of  a  tower  standing  on  level  ground,  the  angles  of  elevation  of  the  top 
are  45°  and  30°,  respectively.     Find  the  height  of  the  tower. 

10.  In  exercise  9,  if  the  angles  of  elevation  are  60°  and  30°,  what  is 
the  height  of  the  tower,  and  how  far  are  the  points  A  and  B  from  its 
base? 

46.  Functions  of  0°,  90°,  180°,  and  270°.  The  functions  of  0°, 
90°,  180°  and  270°  are  not  included  in  the  above  definition  of 
trigonometric  functions,  since  for  these  angles  there  are  no  tri- 
angles of  reference.  We  shall  define  the  functions  of  0°,  90°,  180°, 


88  MATHEMATICS  [§46 

and  270°,  as  the  limits  toward  which  the  functions  of  any  angle, 
say  a,  approach  as  the  angle  a  approaches  0°,  90°,  180°,  or  270°. 
Thus,  if  the  angle  a  is  in  the  first  quadrant,  and  if  the  hypotenuse 
of  the  triangle  of  reference  is  kept  constant  as  the  angle  becomes 
smaller  and  smaller,  the  opposite  side  will  become  smaller  and 
smaller  and  approach  zero.  The  sine  of  this  variable  angle  will 
then  approach  zero  as  the  angle  approaches  zero,  or  the  sin  0°  =  0. 
The  length  of  the  adjacent  side  approaches  the  length  of  the 
hypotenuse  as  the  angle  approaches  zero,  or  the  limit  of  the 
cosine  is  1,  or  cos  0°  =  1.  As  the  angle  approaches  zero,  the 
cotangent  and  the  cosecant  increase  without  limit.1 

By  considering  a  a  small  negative  angle  (hence  in  the  fourth 
quadrant)  the  cotangent  and  cosecant  are  both  negative  and 
increase  numerically  without  limit  as  a  approaches  zero.  These 
facts  together  with  what  was  said  above  regarding  cotangent  and 
cosecant  are  expressed  by  writing  cot  0°  —  +  °°  and  esc  0°  =  +  oo  . 

Thus  we  have : 

sin  0°  =  0,  tan  0°  =  0,  sec  0°  =  1, 

cos  0°  =  1,  cot  0°  =    ±  oo  esc  0°  =  ± 

By  allowing  the  angle  to  approach  90°,  the  student  will  show 
that: 

sin  90°  =  1,  tan  90°  =  ±  oo          sec  90°  =  ±00, 

cos  90°  =  0,  cot  90°  =  0,  esc  90°  =  1. 

1  If  6  is  a  variable  which  approaches  zero  as  its  limit,  and  if  a  is  a  constant  or  a 

a 

variable  whic  approaches  a  limit  c,  the  fraction  -  increases   without  limit.      Thus, 

6 

if  a  =  1,  and  if  b  approaches  zero  by  taking  successively  the  values  1,  1,  J,  J,  .  .  ., 
the  fraction  has  successively  the  values  1,  2,  3,  4,  .  .  .  By  giving  to  6  a  value 

sufficiently  small,  the  value  of  the  fraction,  vt  may  be  made  larger,  and  for  subse- 
quent values  of  6  remains  larger,  than  any  assigned  number.  We  then  say  that 

increases  without  limit  as  6  approaches  zero. 

The  expression,  "r  increases  without  limit,"  or  "increases  indefinitely,"  as  6 
approaches  zero  is  usually  written 

r  =    oo 

The  student  is  not  to  infer  that  °°  (infinity)  is  a  number  or  the  limit  of  a  number- 
The  expression,  r=  oo,is  a  symbol  representing,  "r  increases  without  limit,"  or  "r 
increases  indefinitely." 


CO 


§46]  THE  CIRCULAR  FUNCTIONS  89 

The  student  will  further  show  that: 

sin  180°  =0,  tan  180°  =  0,  sec  180°  =  -1, 

cos  180°  =  -1,       cot  180°  =  ±  co,        esc  180°  =  ±  <», 

and  that: 

sin  270°  =  -1,       tan  270°  =  +  oo ,        sec  270°  =  ±  °° 
cos  270°  =  0,  cot  270°  -  0,  esc  270°  =  - 1 

Exercises 

1.  Tabulate  the  functions  of  0°,  90°,  180°  and  270°. 

2.  Find  the  values  of  the  functions  of  360°;  of  450°;  of  540°;  of 
630°;  of  720°;  of  810°. 

3.  Is  it  possible  for  an  angle  to  have  a  sine  equal  to  1/2?  to  3/4? 
to  -  2? 

4.  Is  it  possible  for  an  angle  to  have  a  cosine  equal  to  1/2? 
to  3/4?  to  -  2? 

5.  Is  it  possible  for  an  angle  to  have  a  tangent  equal  to  1/2? 
to  3/4?  to  -  2? 

6.  Is  it  possible  for  an  angle  to  have  a  cotangent  equal  to  1/2? 
to  3/4?  to  -  2? 

7.  Is  it  possible  for  an  angle  to  have  a  secant  equal  to  1/2?  to  3/4? 
to  -  2? 

8.  Is  it  possible  for  an  angle  to  have  a  cosecant  equal  to  1/2? 
to  3/4?  to  -  2? 

9.  As  the  angle  varies  from  0°  to  360°  in  a  positive  sense,  how  do 
the  values  of  the  sine,  the  cosine,  the  tangent,  the  cotangent,  the 
secant  and  the  cosecant  change? 

10.  In  the  equation  of  the  straight  line,  y  =  ax  +  b,  a  is  what 
trigonometric  function  of  what  angle? 

11.  Compare  the  sin  30°  with  sin  ( —  30°) ;  sin  45°  with  sin  ( —  45°) ; 
sin  60°  with  sin  (-  60°);  sin  135°  with  sin  (-  135°);  sin  300°  with 
sin  (—  300°);  sin  a  with  sin  (—a). 

12.  Compare    cos  30°  with    cos  (- 30°) ;  cos  60°  with    cos  (-60°); 

cos  210°  with  cos  (-  210°);     cos  a  with  cos  (-  a). 

13.  Compare  tan  45°  with  tan  ( -  45°) ;  tan  150°  with  tan  ( -  150°) ; 

tan  330°  with  tan  (-  330°);  tan  a  with  tan  (-a). 

14.  Compare  cos  30°  with  sin  60°;  sin  30°  with  cos  60°;  tan  30° 
with  cot  60°;  sec  30°  with  esc  60°;  sin  45°  with  cos  45°;  tan  45°  with 
cot  45°;  cos  a  with  cos  (90°  —  a);  sin  a  with  cos  (90°  —  a);  tan  a 
with     cot    (90°  —  a);     sec    a    with    esc    (90°  —  a);     esc    a    with 
sec  (90°  —  a),  where  a  is  an  angle  in  the  first  quadrant. 


90  MATHEMATICS  [§47 

16.  Compare  sin  135°  with  sin  45°;  sin  120°  with  sin  60°;  cos  120° 
with  cos  60°;  tan  120°  with  tan  60°;  sin  150°  with  sin  30°;  cos  150° 
with  cos  30°;  tan  150°  with  tan  30°;  sin  240°  with  sin  60°;  cos  240° 
with  cos  60°;  tan  240°  with  tan  60°;  sin  300°  with  sin  60°;  cos  300° 
with  cos  60°;  tan  300°  with  tan  60°;  sin  (180°  -  a)  with  sin  «; 
cos  (180°  -  a)  with  cos  a;  tan  (180°  -  a)  with  tan  «;  sin  (180°  +  a) 
with  sin  a;  cos  (180°  +  a)  with  cos  a;  tan  (180°  +  a)  with  tan  a; 
sin  (360°  —  a)  with  sin  a;  cos  (360°  —  a)  with  cos  a;  tan  (360°  —a) 
with  tan  a. 

16.  Express  the  following  as  functions  of  some  positive  angle  less 
than  90°:  sin  112°;  cos  127°;  tan  171°;  sin  92°;  cos  170°;  tan  131°; 
cot  175°. 

HINT:  Construct  an  angle  of  112°.  Its  terminal  side  falls  in  the 
second  quadrant.  Draw  a  triangle  of  reference.  Construct  an  angle 
of  22°.  Draw  a  triangle  of  reference  for  22°.  These  two  triangles  of 
reference  are  similar,  and  their  homologous  sides  are  in  proportion. 
Whence:  sin  112°  =  +  cos  22°;  cos  112°  =  -  sin  22°  (because  cosine 
is  positive  in  the  first  quadrant,  and  negative  in  the  second),  and 
tan  112°  =  -  cot  22°. 

17.  Show  that 

sin  a 

tan  a  — 

COS  a 

and  that 

cos  a 

cot  a  =  -s • 


47.  Even  and  Odd  Functions.  Let  us  consider  the  function 
x2  (i.e.,  y  =  x2).  If  we  give  to  x  any  two  values  numerically  equal, 
one  positive  and  the  other  negative,  the  two  corresponding  values 
of  x2  (i.e.,  oiy)  are  equal.  The  curve  for  x2  (i.e.,  y  =  x2),  plotted 
upon  squared  paper  is  symmetrical  with  respect  to  the  F-axis. 
The  same  can  be  said  of  the  functions  x4,  x6,  x6,  x2  +  3a;4,  and 
x2  +  5x4  —  6z6  —  7.  These  functions  are  said  to  be  even.  An 
even  function  is  one  whose  value  remains  unchanged  when  the 
sign  of  the  argument  is  changed. 

Let  us  next  consider  the  function  x3  (i.e.,  y  =  x3).  If  we  give  to 
x  two  values  numerically  equal  but  with  opposite  signs,  the  two 
corresponding  values  of  x3  (i.e.,  of  y)  are  numerically  equal  but  of 
opposite  signs.  The  curve  for  x3  (i.e.,  y  =  x3),  plotted  upon 
squared  paper  is  symmetrical  with  respect  to  the  origin.  The 


§48]  THE  CIRCULAR  FUNCTIONS  91 

same  can  be  said  of  the  functions  x5,  re7,  x,  x  +  3x3  —  7x&, 
and  x3  —  2z5  +  6z7.  These  functions  are  said  to  be  odd.  An 
odd  function  is  one  whose  value  'remains  numerically  the  same  but 
changes  sign  when  the  sign  of  the  argument  is  changed. 

In  previous  exercises  the  fact  was  developed  that  cos  a  and  sec  a 
are  even  functions,  and  that  sin  a,  esc  a,  tan  a  and  cot  a  are 
odd  functions.  These  facts  are  shown  graphically  in  §  49. 

48.  Circular  Measure.  The  circular  unit  of  angular  magnitude 
is  defined  as  an  angle  whose  sides  intercept  an  arc  equal  in  length 
to  the  radius  of  a  circle  when  the  vertex  of  the  angle  is  placed  at  the 
center  of  the  circle.  This  unit  is  called  the  radian.1  This  sys- 
tem of  angular  measure  is  fundamental  in  mechanics,  physics,  and 
pure  mathematics.  It  must  be  thoroughly  mastered  by  the 
student. 

Since  the  circumference  of  a  circle  is  2irr,  where  r  is  the  radius  of 
the  circle,  there  are  2ir  radians  in  one  complete  revolution;  or 

360°  =  2-ir  radians 
180°  =  TT  radians 
90°  =  K  radians,  etc. 

6 

To  find  the  number  of  radians  in  an  angle  expressed  in  degrees, 
divide  the  number  of  degrees  by  180  (this  gives  the  number  of 
straight  angles  in  the  given  angle)  and  multiply  the  quotient  by  TT, 
the  number  of  radians  in  a  straight  angle.  Thus,  to  change  176°  to 

1  7  ft 

radians,  we  have  T^TT  =  3.07:  or  176°  =  3.07  radians. 

loU 

To  find  the  number  of  degrees  in  an  angle  expressed  in  radians, 
divide  the  number  of  radians  by  TT  (this  gives  the  number  of 
straight  angles  in  the  given  angle)  and  multiply  this  quotient  by 
180  (the  number  of  degrees  in  a  straight  angle).  Thus,  to  change 

4.6  radians  to  degrees  we  have  — '- =  263. 6,  or  4. 6  radians 

7T 

equals  263.6°. 

1  If  0  is  the  angle  measured  in  radians  at  the  center  of  a  circle  of  radius  r,  the 
length  of  the  arc  intercepted  by  the  angle  is  rO,  and  the  area  of  the  circular  sector 
is  *r20. 


92 


MATHEMATICS 


\ 


\ 


\0 


y 


7 


Exercises 

1 .  Express    in    radians    the 
following:  (a)  135°;  (6)  45°;  (c) 
161°;  (d)  275°. 

2.  Express    in    degrees     the 
following    angles    measured    in 
radians:  (a)   7r/2;    (6)   ir/4;    (c) 
Tr/6;  (d)  37T/2;  (e)  1;  (/)  2;  fo) 
3.6;  (ft)  11.6. 

[49.]  Graphs  of  the  Tri- 
gonometric Functions.  In 
Fig.  26  let  OPiPzPa  be  a 
circle  with  unit  radius.  Let 
OCPi  be  any  angle  at  the 
center  of  the  circle.  Then 
PiAi  represents  the  sine  of 
the  angle  0CP,.  Take  OB 
equal  in  length  to  the  circum- 
ference of  the  circle,  i.e.,  equal 
to  2-n-.  Divide  OB  into  thirty- 
six  equal  parts,  and  erect  per- 
pendiculars at  each  point  of 
division.  Each  of  these  equal 
distances  of  the  X-axis,  OB, 
represents  10°,  orT/18  radian. 
Upon  each  vertical  line  plot  a 
point  whose  ordinate  repre- 
sents the  sine  of  the  corre- 
sponding angle.  To  do  this 
divide  the  circumference  of 
the  circle  into  thirty-six  equal 
parts,  and  join  each  to  C,  the 
center  of  the  circle.  (Only  a 
few  of  these  points,  as  PI,  Pz, 
PZ,  PI,  P&,  are  lettered  in  the 
drawing.)  Project  these 
points  to  the  right  upon  the 
vertical  line  corresponding  to 
the  angle  at  the  center  of  the 


§49] 


THE  CIRCULAR  FUNCTIONS 


93 


circle.  Thus,  the  angle  50°  determines  the  point  P2.  P2  pro- 
jected upon  the  line  passing  through  the  point  of  the  X-axis  repre- 
senting 50°  gives  P'2.  ThenP'2A'2  =  P2A2  =  sin  50°. 

The  curve  OKMNB  (called  the  sine  curve)  is  the  graph  of  the 
equation  y  =  sin  x,  as  x  varies  from  0  to  2?r.  It  is  readily  seen 
that  as  x  continues  beyond  this  interval  the  curve  is  repeated  for 
every  addition  of  2ir  to  the  X-axis. 

The  length  of  any  portion  of  the  X-axis  represents  an  angle  at  the 
center  of  the  circle  expressed  in  circular  units. 

The  graph  of  y  =  2  sin  x  may  be  obtained  from  the  curve  in  Fig. 
26  by  making  all  ordinates  twice  as  long;  the  curve  y  =  3  sin  £  by 
making  all  ordinates  three  times  as  long;  the  curve  y  =  f  sin  x 
by  making  all  the  ordinates  one-half  as  long;  the  curve  y  •=  %  sinx 
by  making  all  ordinates  one-third  as  long;  etc. 


FIG.  27. 


If  the  sine  curve  in  Fig.  27  were  translated  to  the  left  a  distance 
ir/2,  its  equation  would  be  y  =  sin  (x  +  ir/2).  Since  the  sin 
(x  +  7T/2)  =  cos  x,  the  translated  curve  in  Fig.  27  is  the  graph 
for  y  =  cos  x. 

To  construct  the  graph  y  =  tan  x  we  proceed  by  a  method 
similar  to  that  used  in  the  construction  of  the  sine  curve. 

Let  C,  Fig.  28,  be  the  center  of  a  circle  of  unit  radius.  Upon  the 
X-axis  lay  off  the  distance  OD  equal  to  2ir,  the  circumference  of  the 
circle.  Divide  OD  into  thirty-six  equal  parts,  and  at  each  point 
of  division  erect  a  perpendicular.  Let  the  F-axis,  Y'OY,  be 
tangent  to  the  circle.  Beginning  with  the  horizontal  line  draw 
thirty-six  radii  at  intervals  of  10°,  extending  them  to  the  tangent 


94 


MATHEMATICS 


t§50 


line  Y'OY.  Project  the  points  of  intersection  with  the  tangent 
line  upon  the  vertical  line  whose  distance  from  the  origin  repre- 
sents the  particular  angle  at  the  center  of  the  circle.  Thus,  OCPi 
is  an  angle  of  30°.  Project  PI  upon  the  third  vertical  line  to  the 
right  of  the  F-axis,  and  obtain  P'i,  a  point  upon  the  graph  of 
y  =  tan  x.  When  the  angle  is  in  the  second  or  third  quadrants 
the  radius  must  be  extended  back  through  the  center  in  order 
to  intersect  with  the  tangent  Y'OY.  We  see  that  as  the  angle 


FIG.  28. — The  tangent  curve. 

passes  through  90°  and  270°  the  tangent  curve  jumps  from  +  °° 
to  —  «» .  The  tangent  curve  extends  to  infinity  in  both  y  direc- 
tions, and  only  a  portion  of  it  near  the  X-axis  is  represented  in 
the  figure.  Curves  for  y  =  cot  x,  y  =  sec  x,  and  y  =  esc  x  may 
be  plotted  from  the  tangent,  cosine,  and  sine  curves,  respectively, 
by  taking  reciprocals  of  the  lengths  of  the  ordinates. 

[50.]  The  Line  Representation  of  the  Trigonometric  Functions. 
Let  C,  Fig.  29,  be  the  center  of  a  circle  with  unit  radius.  Let  CO 
be  a  horizontal  line  and  CD  a  vertical  line.  Let  OY  and  DQ  be 


[§50 


THE  CIRCULAR  FUNCTIONS 


95 


tangents.  Let  OCP  be  an  angle  a,  at  the  center  of  the  circle. 
Then,  AP  represents  the  sine  of  a;  CA  represents  the  cosine  of  a; 
OP i  represents  the  tangent  of  a;  DP2  represents  the  cotangent  of 
a;  CPi  represents  the  secant  of  a;  and  CPz  represents  the  cose- 
cant of  a. 

AO,  which  is  equal  to  1  —  cos  a,  is  sometimes  called  the  versed 
sine  of  a  (vers  a). 


FIG.  29. 


Exercises 

1.  Draw  a  curve  for  y  =  sin  x.     Use   1.15  inches  as  the  unit 
of  length.     This  gives"  2?r  =  7.2  inches,  approximately,  or  each  0.2 
inch  of  the  X-axis  represents  10°. 

2.  Draw  a  curve  for  y  =  cos  x. 

3.  Draw  a  curve  for  y  =  tan  x. 

4.  Prove  (sin  a)2  +  (cos  a)2  =  1;  or,  as   it  is  usually  written, 
sin2  a  +  cos2  a  =  1,  where  a  is  any  angle. 

HINT:  In  a  triangle  of  reference  for  any  angle,  excluding  0°,  90°, 
180°,  and  270°,  the  square  of  the  adjacent  side  plus  the  square  of  the 
opposite  side  equals  the  square  of  the  hypotenuse.  Note  that  this 
is  true  even  though  we  consider  the  sides  negative.  For  the  angles 


96  MATHEMATICS  [§51 

excepted  above,    prove   the    truth   of    the   formula   by  direct  sub- 
stitution. 

6.  Prove  sec2  a  =  1  +  tan2  a. 

6.  Prove  esc2  a.  =  1  +  cot2  «. 

51.  Trigonometric   Tables.     In   preceding   exercises  we   have 
calculated  the  numerical  values  of  the  trigonometric  functions  of 
a  few  angles;  viz.,  30°,  45°,  60°,  etc.     If  the  values  of  the  functions 
of  other  angles  were  desired  they  could  be  found  by  drawing  the 
angles  with  a  protractor,  constructing  *a  triangle  of  reference,  and 
from  it  scaling  off  the  lengths  of  the  sides  and  computing  the  ratios. 
This  method  of  finding  the  functions  would  not  only  be  slow,  but 
the   degree   of   accuracy   would   usually   be   unsatisfactory.     A 
method  has  been  devised  for  calculating  the  trigonometric  func- 
tions to  any  number  of  decimal  places,  but  it  cannot  be  explained 
in  this  place. 

A  table  of  natural  trigonometric  functions,  for  every  ten  minutes 
from  0°  to  90°,  is  given  in  Table  XXII,  pages  300-302.  Since  the 
cosecant  and  secant  are,  respectively,  reciprocals  of  the  sine  and 
cosine,  they  need  not  be  used  in  formulas,  and  their  values  are  not 
usually  published  in  a  trigonometric  table.  If  the  angle  is  less 
than  45°  the  degrees  and  minutes  of  the  angle  are  printed  in  the 
left-hand  column  and  the  names  of  the  functions  are  printed  at  the 
top  of  the  page.  If  the  angle  is  greater  than  45°  the  degrees  and 
minutes  are  printed  in  the  right-hand  column  and  the  names  of 
the  functions  are  printed  at  the  bottom  of  the  page.  This 
arrangement  of  the  table  is  possible  through  the  fact  that  the 
cosine  of  an  angle  is  the  sine  of  its  complement,  and  the  cotan- 
gent is  the  tangent  of  its  complement. 

52.  Graphic    Table   of   Trigonometric    Functions.     From    a 
sheet  of  polar  coordinate  paper,  form  MS,  the  sine  and  the  cosine  of 
any  angle,  and  the  tangent  of  an  angle  less  than  45°,  may  be  scaled 
off  accurately  to  within  two  or  three  points  in  the  third  decimal 
place. 

Upon  this  sheet  concentric  circles,  2/100  of  a  unit  apart,  are 
drawn.  Every  fifth  circle  is  shown  by  a  heavy  line.  Running 
from  their  common  center  radiating  lines  for  every  degree  are 
drawn.  Every  tenth  line  is  drawn  heavy.  Two  additional  circles 
are  drawn  each  passing  through  the  origin,  the  center  of  the  con- 


§52] 


THE  CIRCULAR  FUNCTIONS 


97 


centric  circles,  and  each  having  a  diameter  equal  to  unity.  The 
circle  with  its  center  upon  the  F-axis  is  called  the  sine  circle,  the 
other  with  its  center  upon  the  X-axis  is  called  the  cosine  circle. 
At  the  right  is  a  vertical  scale  from  which  may  be  read  the  tan- 
gent of  any  angle  up  to  45°.  This  scale  is  called  the  tangent  scale, 
and  its  use  is  apparent  at  once.  In  explaining  the  method  of 
scaling  off  the  sines  and  cosines  from  the  sine  and  cosine  circles, 


FIG.  30. 


reference  is  made  to  Fig.  30.  Let  0  be  the  center  of  the  system 
of  concentric  circles,  and  let  CABD  be  the  circle  with  radius  equal 
to  unity,  and  center  at  0.  Let  OEA  be  the  cosine  circle,  and  OFB 
the  sine  circle.  Let  AOE  be  any  angle,  say  a.  Draw  AE  and 
FB.  OAE  and  OFB  are  right  triangles.  Then  OE  =  OA  cos  a. 
But  OA  =  1.  Therefore  OE  =  cos  a.  Upon  form  MS  the 
length  OD  may  be  read  off  at  once  by  means  of  the  system  of 
concentric  circles. 

Similarly,  OF  =  OB  sin  OBF.    But,  since  OB  =  1  and  the  angle 
OBF  =  angle  a,  we  have  OF  =  sin  a. 

7 


98  MATHEMATICS  [§53 

63.  The  Graphic  Solution  of  the  Right  Triangle.    Let  A,  B, 

and  C  be  the  vertices  of  any  triangle;  and  let  a,  /3,  and  y  be,  re- 
spectively, the  interior  angles  of  the  triangle  at  these  vertices.  Let 
the  sides  opposite  the  vertices  A,  B,  and  C  be,  respectively,  a,  b, 
and  c.  In  a  right  triangle  the  right  angle  is  marked  7,  and  the 
hypotenuse  c. 

The  three  sides  together  with  the  three  angles  are  called  the 
elements  of  the  triangle.  If  any  three  of  the  six  elements,  except- 
ing the  three  angles,  of  a  possible  triangle  are  given,  the  triangle 
may  be  constructed  to  scale,  and  the  remaining  three  elements 
measured.  The  triangle  is  then  said  to  be  solved  graphically. 

Exercises 

Solve  graphically  the  following  right  triangles;  record  the  time  spent 
upon  each  problem. 

l.o  =  173,     b  =  216.  4.  a  =  175,     a  =  36°. 

2.  b  =  136,     c  =  527.  5.  c  =  516,     0  =  73°. 

3.  a  =  210,     c  =  728.  6.  b  =  172,     a  =  47°. 

54.  Solution  of  Right  Triangles  Analytically.  Instead  of  solv- 
ing a  right  triangle  graphically  we  may  solve  it  analytically. 
This  method  will  be  illustrated  by  solving  two  right  triangles. 

ILLUSTRATION  1 :  Given,  a  =  216  and  a  =  36°;  to  find  b,  c,  and 
/3.  We  see  at  once  that  /3  =  54°,  the  complement  of  36°.  To  find  c 

a,  a 

use  the  relation  c  =  —  — ,  and  to  find  b  use  the  relation  b  =  —    — 
sin  a  tan  a 

From  the  table  of  the  natural  trigonometric  functions, 
sin  a  =  0.5878,  and  tan  a  =  0.7265. 

216  216 

Substituting,  c  =  --—  =  367.5;  and  b  =  ~—  -  297.3. 

To  check  the  result,  make  use  of  the  relation 

c2  -  a2  =  62, 
or  \/(c  —  a)(c  +  a)  =  b. 

c  -  a  =  151.4. 
c  +  a  =  583.4. 
V'(c^-~aXc  +"a)  =  297.3  =  b 


§54]  THE  CIRCULAR  FUNCTIONS  99 

This  last  value  of  6  agrees  with  the  value  computed  above. 
This  not  only  checks  the  value  of  b,  but  also  the  value  of  a,  which 
is  used  in  computing  the  second  value  of  6. 

ILLUSTRATION  2:  Given  c  =  516.2  and  a  =  176.5;  to  find  b,  a, 
andjS. 

a       176.5 

Solution:      sin  a  =  -  =  r<  .  _  =  0.3419 
c       51o.2 

From  the  table  a  =  19°.994 

whence  0  =  70°.006 

b  =  csin/3  =  (516.2)  (0.9397)  =  485.1 


Check:  \/(c  —  a)(c  +  a)  =  b, 

or  \/235,310.19         =  6, 

or  485.1  =  6, 

which  checks  all  work. 

Exercises 

Solve  the  following  right  triangles.  Use  natural  values  of  trigono- 
metric functions  taken  from  Table  XXII.  Perform  the  necessary 
multiplications  and  divisions  by  the  arithmetical  process,  and  hand 
in  all  numerical  work.  Be  sure  to  check  your  work.  Keep  a  record  of 
time  spent  upon  each  problem. 

[1.]  a  =  173,  b  =  216.  [4.]  a  =  175,  a  =  36°. 

[2.]  b  =  136,  c  =  527.  [5.]  c  =  516,  ft  =  73°. 

[3.]  a  =  210,  c  =  728.  [6.]  b  =  172,  a  =  47°. 

Instead  of  performing  the  multiplications  and  divisions  by  the 
lengthy  arithmetical  method,  logarithms  may  be  used.  In  the 
logarithmic  solutions  the  natural  values  of  the  trigonometric  func- 
tions are  not  used,  because  their  logarithms  may  be  taken  directly 
from  a  logarithmic  table.  The  two  problems  solved  below  illus- 
trate the  arrangement  of  the  work.  The  student  should  go  over 
the  solutions  and  checks  of  these  two  problems  very  carefully,  in 
order  to  become  familiar  with  the  use  of  a  table  of  the  logarithms 
of  the  trigonometric  functions.  Especial  study  should  be  given 
to  the  finding  of  the  angles  when  the  logarithms  of  the  functions 
are  given. 


100  MATHEMATICS  [§54 

ILLUSTRATION  1:  Given  a  =  175.6  and  6  =  216.1;  to  find  a 
/?,  and  c. 

a  6 

Solution:      tan  a  =  ~ ,      /3  =  180  —  a,  c  =  — — - 
6  sm  /S 

log  a  =  2.2443 
log  6  =  2.3347 
log  tan  a  =  9.9096  -  10 
a  =  39°  4' .6 
0  =  50°  55' .4 
log  sin  ]8  =  9.8900  -  10 
log  c  =  2 . 4447 
c  =  278.44 
Check: 

c  -  a  =  102.84 

c  +  a  =  454.04 

log  (c  -  a)  =  2.0122 

log(c  +  a)  =  2.6571 

logfc2  =  4.6693 

log  6  =  2.3347 

The  computed  log  6  agrees  with  log  b  used  above,  which  checks 
all  of  the  work. 

ILLUSTRATION  2:  Given,  a  =  176.3,  a  =  37°  17'.4;  to  find  b,  c, 
and  j9. 

Solution:   /3  (the  complement  of  a)  =  52°  42'. 6.     b  =  a  tan  /3, 

a 

and  c  =  — — -• 
smp 

log  a  =  2.2463 

log  tan  /3  =  0.1183 

logfe  =  2.3646 

6  =  231.53 

log  sin  |8  =  9.7824  -  10 
logc  =  2.4639 
c  =  291 . 00 
Check: 

c-  a  =  114.7 
c  +  a  =467.3 
log  (c  -  a)  =  2,0596 


§54]  THE  CIRCULAR  FUNCTIONS  101 

log  (c  +  o)  =  2.6696 
log&2  =  4.7292 
log  6  =  2.3646 

which  agrees  with  log  b  above  and  thus  checks  the  entire  work. 

This  illustrative  example  was  chosen  so  that  the  calculated  value 
of  the  logarithm  of  b  would  agree  with  the  value  found  in  the  table. 
This  absolute  agreement  will,  in  general,  not  occur.  For,  in 
finding  the  second  logarithm  of  b,  calculated  numbers  are  used 
which  are  correct  only  to  four  places,  and  the  errors  in  these 
numbers,  especially  in  c-a  will  effect  the  calculated  logarithm 
of  6. 

If  the  calculated  logarithm  differs  from  that  taken  from  the 
table  by  only  one  or  two  in  the  fourth  decimal  place,  the  student 
may  consider  his  work  checked. 

In  checking  the  work  it  will  be  noticed  that  it  is  immaterial 
whether  the  log  a  or  log  b  is  recomputed.  If  the  hypotenuse  and  a 
leg  are  given,  the  logarithm  of  the  unknown  leg  should  be  recom- 
puted. Otherwise  the  logarithm  of  the  longer  leg  should  be 
recomputed  by  the  check  formula.  The  reason  for  this  is  that 
it  keeps  the  percent  of  error  in  the  difference  between  the 
hypotenuse  and  the  leg  as  small  as  possible. 

Exercises 

Solve  the  following  right  triangles  using  logarithms.  Keep  a  record 
of  the  time  spent  upon  each  problem.  Check  the  work. 

l.o  =  173,  b  =  216.  4.  a  =  175,  a  =  36°. 

2.  6  =  136,  c  =  527.  5.   c  =  516,  ft  =  73°. 

3.  a  =  210,  c  =  728.  6.  b  =  172,  a  =  47°. 

7.  Having  measured  a  distance  of  281.6  feet  in  a  direct  horizontal 
line  from  the  bottom  of  a  tower,  the  angle  of  elevation  of  the  top  was 
found  to  be  24°  16'.6.     Find  the  height  of  the  tower. 

8.  A  person  on  top  of  a  tower  75  feet  high  observes  the  angle  of 
depression  of  two  objects  on  the  horizontal  plane,  which  are  in  line 
with  the  base  of  the  tower,  to  be  37°  27'.6  and  25°  17'.8.      Find   the 
distance  of  each  object  from  the  base  of  the  tower. 

9.  A  tower  stands  by  a  river.     A  person  on  the  opposite  bank 
finds  the  angle  of  elevation  of  its  top  to  be  50°.     He  recedes  40  yards 


102 


MATHEMATICS 


[§54 


in  a  direct  line  from  the  tower,  and  finds  the  angle  of  elevation  of  the 
top  to  be  40°.  Find  the  breadth  of  the  river  and  the  height  of  the 
tower. 

10.  A  rope  100  feet  long  is  fastened  to  the  top  of  a  building  40  feet 
high.     Find  the  angle  the  rope  is  inclined,  if  its  lower  end  just  touches 
the  ground. 

11.  From  a  balloon,  which  is  directly  over  one  town,  is  observed 
the  angle  of  depression  of  another  town,  12°  16'.     The  towns  are 
8.5  miles  apart.     Find  the 

height  of  the  balloon. 

13.  From  a  station  A  at 
the  base  of  a  mountain,  its 
summit  S  is  seen  at  an  ele- 
vation of  43°  17'.  After 
walking  5000  feet  toward 
the  summit,  up  a  plane 
making  an  angle  of  28°  13' 
with  the  horizontal,  to 
another  station,  B,  the  angle 
ABS  was  found  to  be  137° 
17'.  Find  the  height  of 
the  mountain. 

HINT:  In  solving  any  problem,  first  build  up  a  formula  giving 
the  unknown  number.  As  far  as  possible  put  this  formula  in  a  form 
suitable  for  logarithmic  computation.  This  will  be  illustrated  by 
building  up  a  formula  for  x,  the  height  of  the  mountain.  In  Fig.  31, 
SE  =  x  is  the  required  height.  BA  =  5000  feet;  a  =  28°  13'  and 
the  angles  /3,  y,  and  5  can  be  computed  from  the  given  data.  BC  is 
perpendicular  to  AS. 

BC  =  5000  sin  /3 

BC        5000  sin  ft 


BS=    . 

sin  5  sin  8 

DS  =  BS  sin  y 

=  5000  sin  0  -7 

sin  5 

DE  =  BF  =  5000  sin  a 

SUIT 

x  =  DS  +  DE  =  5000  sin  /3  - 

sin  5 


5000 


sin  S 


+  5000  sin  a 
+  sin  a 


THE  CIRCULAR  FUNCTIONS 


103 


Compute  /3,  y,  and  5,  and  substitute  in  the  formula.  Compute  the 
value  of  the  first  term  within  the  parenthesis  by  using  logarithms. 

13.  From  a  point  in  a  window  in  the  same  horizontal  plane  with  the 
bottom  of  a  steeple,  the  angle  of  elevation  of  the  top  of  the  steeple  is 
43°  16'.  From  another  window,  18  feet  below  the  first,  the  angle 
of  elevation  of  the  top  of  the  steeple  is  52°  8'.  Build  up  a  formula 
giving  the  height  of  the  steeple.  Find  the  height. 


A  c  B          A  c  B 

FIG.  32. 

55.  The  Law  of  Sines.    Let  ABC,  Fig.  32,  be  any  triangle.    Let 
p  be  a  perpendicular  let  fall  from  C  upon  c.     Then 

p  =  a  sin  /3 
p  =  b  sin  a. 
a  sin  j8  =  b  sin  a, 

a         _b_ 

sin  a  ~  sin  £ 

In  a  similar  way  we  may  show  that 


or 


a 


Hence 


sin  a      sin  /3       sin  7 
which  is  known  as  the  law  of  sines. 

Exercises 

1.  Suppose  we  are  required  to  determine  the  distance  from  a 
given  point  A  to  an  inaccessible  object  B,  Fig.  33.    Let  C  be  any  other 


104 


MATHEMATICS 


[§55 


convenient  accessible  point.  Set  a  transit  over  A  and  measure  the 
angle  B AC.  Set  the  transit  over  C  and  measure  the  angle  BCA. 
Measure  the  distance  AC.  Solve  for  AB,  using  the  law  of  sines. 


Data: 


b  =  500  feet 
a  =  101°  13'  6 

7  =    42°28'.7 


FIG.  33. 


To  find  c  =  AB,  use  the  formula 


sin  7       sin  /3 
or 

b  sin  7 
sin  /3 

The  value  of  /3  may  be  found  by  subtracting  the  sum  of  a  and  7 
from  180°,  for,  from  geometry,  the  sum  of  the  three  interior  angles  of 
a  triangle  is  equal  to  180°. 

2.  Same  as  exercise  1,  with  the  following  data: 

b  =  375.6  feet 
a  =  27°  42'.  7 
7  =  113°  17'.6. 

HINT:     Sin  (113°  17'.6)  =  sin  (66°  4.2'  A.)  =  cos  (23°  17'.6) 


§56] 


THE  CIRCULAR  FUNCTIONS 


105 


56.  The  Law  of  Cosines.    Let  ABC,  Fig.  34,  be  any  triangle. 
Let  p  be  a  perpendicular  let  fall  from  B  upon  6. 

C2  =  p2  +  (6  -  k)* 
if  7  is  acute  (Fig.  34,  a). 


Since  pz  +  #2  =  a2,  we  have, 

C2  =  02  +  52 

and  since  k  =  a  cos  7,  we  have 


62  —  2a&  cos  7. 


FIG.  34. 

If  7  is  obtuse  (Fig.  34,  6), 

c2  =  p2  +  (6  +  A;)2 

c2  =  p2  +  62  +  26fc  +  A;2 

c2  =  a2  +  62  +  2kb: 

But,  since  k  =  a  cos  D(7#  =  —  a  cos  7,  we  obtain  the  same  expres- 
sion for  c  as  above, 

c2  =  a2  +  b2  -  2ab  cos  7, 
which  is  known  as  the  law  of  cosines. 

Exercises 

1.  Suppose  we  wish  to  determine  the  distance  between  two  points 
A  and  B  on  opposite  sides  of  a  building,  Fig.  35.     Choose  any  con- 
venient point,  C,  and  measure  AC,  CB,  and  the  angle  ACS. 
Data  : 

a  =  137.6  feet 
b  =  186.5  feet 
7  =  76°  54'.6. 
To  find  c  use  formula 

C2  =  a2  +  b2  -  2abcos7. 


106 


MATHEMATICS 


[§57 


2.  Same  as  exercise  1,  with  the  following  data: 
a  =  378.6  feet. 
b  =  276.5  feet. 
7  =  112°27'.8. 


57.  The  Law  of  Tangents.  Let  ABC,  Fig.  36,  be  any  triangle. 
With  C  as  center  and  with  CA  as  radius  describe  an  arc  cutting  BC 
at  E  and  BC  produced  at  D.  Draw  AE  and  AD.  BAD  is  a  right 


triangle  with  the  angle  AEC  equal  to 


The  angle  EAB  is 


equal  to 


Draw  EK  perpendicular  to  AE. 


K 


FIG.  36. 


The  triangles  BEK  an4  BAD  are  similar.     Hence 


AD  AE  a  + 

-  =  tan  — 
AE  EK  2 


BD  _  AD 
BE  ~  EK 
Since  BD  =  a  +  b,  and  BE  =  a  —  b, 

a  +  b  _  tan  J  (a  + 
a  —  b       tan  \  (a  — 


cot 


(D 


§57]  THE  CIRCULAR  FUNCTIONS  107 

which  is  known  as  the  law  of  tangents.     Since  a  +  /3  +  7  =  180°, 


2  2' 

The  law  of  tangents  may  be  used  instead  of  the  law  of  cosines 
for  solving  triangles  in  which  two  sides  and  the  included  angle  are 
given. 

ILLUSTRATION  :    Given 

a  =  337.  6  feet 
b  =  213.  5  feet 
7  =  27°  16' 
to  find  c,  a  and  |8. 

Solution: 

tan  $(«  -  |8)  =  CI—  ^-  tan  \(a  +  j8) 
a  +  6 

a  -  6  =  124.1 
a  +  b  =  551.1 
J(a  +  j8)  =  76°  22' 
log  (a  -  6)  =  2.0938 
log  tan  |(a  +  0)  =  0.6152 
log  (o  +  6)  =  2.7413 
log  tan  |(a  -  0)  =  9.9677  -  10 
i(«  _  0)  =  42°  52' 
a  =  119°  14' 
/8  =  33°  30' 

Find  the  value  of  c  by  using  the  law  of  sines, 

b  sin  7 
c  =  — 

sin  |8 

log  6  =  2.3294 
log  sin  7  =  9.6610  -  10 
log  sin  /3  =  9.7419  -  10 

log  c  =  2.2485 
c  =  177.2 


108  MATHEMATICS  [§58 

Exercises 
Solve  exercises  1  and  2,  §  66,  by  using  the  law  of  tangents. 

58.  The  Addition  Formulas.  Let  ABC,  Fig.  37,  be  any  tri- 
angle. Let  ACD  be  the  exterior  angle  formed  by  producing  the 
side  BC.  Draw  CK  perpendicular  to  AB. 


The  law  of  sines  gives, 

AB  sin  a 
smy  =    ~BC—  (1> 

Since  a  +  ft  =  180°  —  7,  sin  (a  +  ft)  =  sin  7.     Equation  (1)  may 
then  be  written 

.     ,          _.       AB  sin  a 
sm  (a  +  ft)  =  —      — 


or 

sin  (a  +  /3)  =  sin  a  cos  /?  +  cos  a  sin  ft.  (2) 

This  formula  gives  the  sine  of  the  sum  of  two  angles,  provided  the 
sum  is  less  than  180°.  We  shall  show  that  this  limitation  may, 
however,  be  removed.  Let  a  +  ft  be  greater  than  180°  but  less 
than  360°.  We  then  have  a  +  ft  =  180°  +  €  +  ft,  where  e  +  ft 
is  less  than  180°. 

sin  (a  +  ft)  =  sin  [180°  +  (  e+  ft)]  =  -sin  (e  +  ft) 
By  formula  (2) 

sin  (e  +  ft)  =  sin  e  cos  ft  -f-  cos  e  sin  /3 


§58]  THE  CIRCULAR  FUNCTIONS  109 

and  since  e  =  a  —  180°, 

sin  (a  +  ft)  =  -  [sin  (a  -  180°)  cos/3  +  cos  (a  -  180°)  sin  ft], 
or 

sin  (  a  +  j8)  =  sin  a  cos  /3  +  cos  a  sin  /3, 
which  shows  that  formula  (2)  is  true  if  a.  +  ft  is  less  than  360°. 

Similarly  it  may  be  shown  that  formula  (2)  is  true  for  any  value 
of  the  sum,  a  +  ft- 

To  prove  a  similar    formula  for  sin  (8  —  a),  drop  a  perpen- 
dicular, BM,  Fig.  37,  from  B  upon  AC: 

sin  ft  =  sin  (5  —  a) 
By  the  law  of  sines 

s*n  a 


— 
n/*  = 


BC 

AM  +  MC    . 
_^_-Bin« 

BMAM   ,   MC  . 


or 

sin  (5  —  a)  =  sin  7  cos  a  +  cos  7  sin  a  (3) 

Since  sin  7  =  sin  5  and  cos  7  =  —  cos  5,  equation  (3)  becomes 

sin  (d  —  a)  =  sin  5  cos  a  —  cos  6  sin  a  (4) 

Formula  (4)  gives  the  sine  of  the  difference  of  two  angles  pro- 

vided neither  angle  exceeds  180°.     The  student  will  show  that  this 

limitation  can  be  removed  in  a  way  similar  to  the  way  in  which  it 

was  removed  from  formula  (2)  above. 

Formulas  (2)  and  (4)  are  usually  written  together  as, 

sin  (a  ±  ft)  =  sin  a  cos  ft  ±  cos  as  in  /3.  (5) 

Since  cos  (a  ±  ft)  =  sin  [(90°  -  a)  +  ft], 
formula  (5)  gives 

cos  (a  ±  ft)  =  sin  (90°  -  a)  cos  ft  +  cos  (90°  -  a)  sin  ft, 
or 

cos  (a  ±  ft)  =  cos  a  cos  ft  +  sin  a  sin  /3.  (6) 

Formulas  (5)  and  (6)  are  called  the  addition  formulas  for  the 
sine  and  cosine. 
If  ft  =  a,  formula  (6)  gives 

cos  («+«)  =  cos  a  cos  a  —  sin  a  sin  a, 


110  MATHEMATICS  [§59 

or 

cos  2  a  =  cos2  a  —  sin2  a,  (7) 

or 

cos  2a  =  1  —  2  sin2  a,  (8) 

since  cos2  a  =  1  —  sin2  a. 

cos  2a  =  2  cos2  a  —  1  (9) 

When  /?  =  a,  formula  (2)  becomes 

sin  (a  +  a)  =  sin  a  cos  a  +  cos  a  sin  a 

sin  2a  =  2  sin  a  cos  a.  (10) 

Exercises 

1.  Find  without   using  the   tables,  the   exact  value  of   sin  75°, 
and  of  cos  75°. 

HINT:  75°  =  45°  +  30°. 

2.  Find  without  using  the  tables,  the  exact  value  of  sin  15°; 
and  of  cos  15°. 

HINT:  15°  =  45°  -  30°. 

X  I  1     —   COS  X 

3.  Show  that  sin—  =  -\ • 

HINT:  Let  2  a  =  x  in  formula  (8). 


4.  Show  that  cos  ~  =  \  I  — 

Zi 

HINT:  Let  2  a  =  x  in  formula  (9). 
6.  Show  that 


x  11  —  cos  x 

tan  -  =  \  — 

2        \  1  +  cos  x 


1  —  cos  x 


sin  x 
sin  x 

1  +  cos  x 
HINT:  Use  results  of  exercises  4  and  5. 

[59.]  Area  of  Triangle  by  Drawing  to   Scale.      In  §  41,  the 

areas  of  triangles,  with  the  three  sides  given,  were  found  both 
graphically  and  analytically. 


§60]  THE  CIRCULAR  FUNCTIONS  111 

In  the  following  exercises  construct  the  triangle  to  scale.  Drop 
a  perpendicular  from  any  vertex  upon  the  opposite  side.  Measure 
the  length  of  this  perpendicular,  and  multiply  this  length  by  the 
length  of  the  side  upon  which  it  was  dropped.  One-half  of  this  prod- 
uct is  the  area  of  the  triangle.  In  drawing  the  triangle  use  a 
convenient  scale,  just  small  enough  to  admit  drawing  upon  a  sheet 
of  paper,  8J  X  11  inches.  Always  mark  in  words  upon  the  draw- 
ing the  scale  used,  as  "scale  1  inch  =  1  foot";  "scale  1/2  inch 
=  10  rods";  "scale  1/4  inch  =  20  rods,"  etc. 

Exercises 

Find  the  area  of  each  of  the  following  triangles.  Record  time  spent 
upon  each  problem.  Place  but  one  problem  upon  a  sheet  of  paper. 

l.o  =  175.6  rods;  b  =  276.5  rods;  /3  =  27°  35'. 

2.  a  =  267.75  feet;  ft  =  37°  45';  7  =  43°  17'.6. 

3.  b  =  262.5  feet;  a  =  178.4  feet;  /3  =  34°  47'.5. 

4.  b  =  175.37  rods;  c  =  216.82  rods;  a  =  27°  47'. 

6.  b  =  73.78  feet;          c  =  53.62  feet;  a  =  136°  43'.6. 

6.  b  =  100.37  feet;        a.  =  27°  31'.6;  ft  =  73°  6'.7. 

60.  Area  of  Triangle  by  Formula.  Case  I.  Two  sides  and  the 
included  angle  given. 

Let  ABC  be  any  triangle  with  b,  c  and  a  given.  From  C  draw 
p  perpendicular  to  the  side  AB.  Then 

area  =  %pc. 
Since  p  =  b  sin  a, 

area  =  i&c  sin  a. 

ILLUSTRATIONS:  Find  the  area  of  the  triangle  having  given 

b  =  172. 36  feet 
c  =  103. 27  feet 
a  =  27°  17'.6. 
log  b  =  2.2364 
logc  =  2.0140 
log  sin  a  =  9.6614  -  10 
log  £  =  9.6990  -  10 
log  area  =  3.6108 

area  =  4081  square  feet 


112  MATHEMATICS  [§60 

Exercises 

Find  the  area  of  each  of  the  following  triangles: 
l.o  =  173.6  rods  b  =  263.8  rods  y  =  73°  47'.6. 

2.  a  =  278.3  rods  c  =  172.7  rods  0  =  21°  21'.7. 

3.  6  =  100.1  rods  c  =  216.5  rods  a  =  127°  26'.3. 
Case  II:    One  side  and  two  angles  given. 

If  two  angles  are  given  the  third  can  be  computed  by  sub- 
tracting their  sum  from  180°.     Let  ABC  be  any  triangle  with 
c,  a  and  /3  given.     From  C  drop  a  perpendicular,  p,  upon  the 
side  c.    Let  M  be  the  foot  of  this  perpendicular. 
Then 

AM  =  p  cot  a 
BM  =  p  cot  ft 
and  by  adding 

c  =  p(cot  a  +  cot  ft) 
'cos  a        cos  j8"j 
.  sin  a         sin  0  J 


=  P 

=  P 
=  P 


sin  ft  cos  a  +  cos  ft  sin 


sin  a  sin  ft 
sin  (a  +  0) 


in  al 


sin  a  sin  /3 

Then 

_  c  sin  a  sm  p 

P  ~  sin  (a  +  0)  ' 
Since  area,  A  =  %pc, 

_  c2  sin  a  sin  / 
= 


2sin(a:+p') 

ILLUSTRATION:    Find    the  area  of  the  triangle  having  given 
c  =  263.75  feet,  a  =  33°  17'.  Q,  and  ft  =  73°  21'.6. 
a  +  /3  =  106°  39'.  2 
logc  =  2.4212 
logc2  =  4.8424 
log  sin  a  =  9.7395  -  10 
log  sin  ft  =  9.9814  -  10 
log  J  =  9.6990  -  10 
log  sin  (a  +  ft)    =  9  .  9816  -  10 
log  area  =  4  .  2807 

area  =  1909  square  feet 


§60]  THE  CIRCULAR  FUNCTIONS  113 

The  log  sin  (a  +  /3),  or  the  log  cos  16°  39'.2,  is  subtracted 
from  the-  sum  of  the  four  preceding  logarithms.  The  subtrac- 
tion is  performed  digit  by  digit,  as  the  addition  progresses 
from  right  to  left. 

Exercises 

Find  the  area  of  each  of  the  following  triangles.  Record  time 
spent  upon  each  exercise. 

1.  c  =  216.3  feet;  «  =  24°  16'.7;  ft  =  61°  15'.6. 

2.  a  =  372.6  feet;  a  =  37°  27'.7;  /3  =  81°  21'.6. 

3.  b  =  87.75  feet;  ft  =  16°  37'.6;  7  =  63°  21'.7. 

4.  c  =  217.75  feet;  a  =  93°  17'.5;  0  =  21°  37'.6. 

5.  a  =  126°  17'.6;  7  =  25°  57'.3;  b  =  200.37  feet. 

[Case  III:]    Given  two  sides  and  an  angle  opposite  one  of  them. 
Let  ABC  be  any  triangle.    Let  a,  c,  and  a  be  given.     Drop  a 
perpendicular,  p,  from  B  upon  AC.    Let  M  be  the  foot  of  the 
perpendicular. 

MA  =  c  cos  a, 
p   =   c  sin  a 

MC  =  Va?  -  p2  =  A/a2  -  c2  sin2~a 
AC  =  AM  ±  MC 

The  minus  sign  is  used  if  7  is  obtuse.     Substituting  for  AM  and 
MC,  ^_____ 

AC  =  c  cos  a  ±  \/a2  —  c2  sin2  a. 
Then,  since  area  =  %pA  C, 

area  =  fc  sin  a  [c  cos  a  ±  A/a2  —  c2  sin2  a] 
area  =  |c2  sin  a  [cos a  ±  \/(a/c  —  sin  a)(a/c  +  sin  a)] 
If  a  /c  <  sin  a,  the  quantity  under  the  radical  sign  is  negative, 
which  means  that  in  this  problem  there  is  no  real  area.     From 
a  figure  it  will  be  seen  that  this  is  the  impossible  case  studied  in 
geometry. 

If  a/c  =  sin  a  we  have  a  right  triangle,  and  the  formula  reduces 
to:  area  =  ^c2  sin  a  cos  a,  which  is  as  it  should  be. 

If  7  is  acute  use  the  +  sign  before  the  radical;  if  obtuse  use  the 
-  sign. 

If  a  >  c,  and  if  a  is  the  interior  angle  of  the  triangle,  there  is  but 
one  solution  given  by  the  formula  with  the  +  sign  before  the 
radical. 


114  MATHEMATICS  [§61 

ILLUSTKATION:     Find  the  area  of  the'triangle  if  c  =  216.7  feet, 
a  =  187.56  feet,  and  a  =  37°  15'.6,  and  7  obtuse, 
log  a  =  2 . 2730 
logc  =  2.3359 
loga/c  =  9.9371  -  10 

a/c  =  0.8652 

sin  a  =  0.6054 

a/c  —  sin  a  =  0.2598 

a  /c  +  sin  a  =  1 . 4706 

log(a/c  -  sin  a)  =  9.4147  -  10 

log  (a/c  +  sin  a)  =0. 1675 

sum  =  9.5822  -  10 

£sum  =  9.7911  -  10 

\/(a/c)2  -  sin2  a  =  0.6181 

cos  a  =  0.79581 

cos  a  —  \/(a/c)2  —  sin2  a  =  0. 1777 

log  0.1777  =  9.2497-  10 
log  sin  a  =  9.7821  -  10 
log  c2  =  4.6718 
log£  =  9.6990  -  10 
log  area  =  3 . 4026 

area  =  2527  square  feet. 

Exercises 

Find  the  areas  in  acres,  of  each  of  the  following  triangles.  Record 
time  spent  upon  each  problem. 

1.  c  =  317.62  rods,  a  =  217.75  rods,  a  =  27°  41'.7,  and  y  is  obtuse. 

2.  c  =  267.7  rods,   a  =  298.6  rods,  a  =    87°  8'. 6,  and  y  is  acute. 
Case  IV.     Three  sides  given.     This  case  has  already  been  dis- 
cussed in  §  41. 

61.  Graphic  Method  of  Constructing  the  Curve  of  Intersection 
of  a  Right  Circular  Cylinder  with  a  Plane.  In  exercise  3,  §  42, 
there  was  described  a  method  of  constructing  the  curve  of  inter- 
section of  a  right  circular  cylinder  and  a  plane  making  an  angle  of 
60°  with  the  axis  of  the  cylinder.  A  graphic  method  will  now  be 

1  In  taking  out  the  cosine  or  the  logarithm  of  the  cosine  of  an  sngle,  the  correc- 
tion, if  any,  must  be  subtracted,  because,  in  the  first  quadrant,  the  cosine  decreases 
as  the  angle  increases. 


§61] 


THE  CIRCULAR  FUNCTIONS 


115 


described  for  the  construction  of  this  curve,  not  only  when  the 
angle  is  60°,  but  for  any  given  angle. 

Let  BQCB',  Fig.  47,  be  the  normal  cross  section  of  the  cylinder,1 
and  let  BPAB'  be  the  section  made  by  a  plane  making  an  angle 
a  with  the  normal  cutting  plane.  From  the  explanation  given  in 
exercise  3,  §  42,  it  is  readily  seen  that 


RP  = 


RQ 
cos  a 


=  RQ  sec  a. 


Draw  the  circle  BQCB',  Fig.  38,  with  radius  OB  equal  to  the 
radius  of  the  cylinder.  Extend  the  diameter  BOB'  to  some  con- 
venient point  0'.  Consider  O'B  and  O'Y'  a  set  of  rectangular 


R     B 


FIG.  38. 


coordinate  axes.  Through  0'  draw  the  45°  line  O'K  and  the  line 
O'M  whose  slope  is  equal  to  the  value  of  sec  a.  Draw  QN  par- 
allel to  O'B  intersecting  O'K  at  N.  Draw  NS  parallel  to  O'Y' 
intersecting  O'M  at  S.  Draw  SP  parallel  to  O'B  intersecting  RQ, 
produced,  at  P.  P  is  a  point  on  the  curve  to  be  drawn.  For, 
since  O'E  =  EN  =  RQ,  and  since  ES  =  RP  =  O'E  sec  a, 
RP  =  RQ  sec  a. 

In  this  manner  any  number  of  points  may  be  located  upon  the 
curve. 

In  the  above  construction  the  horizontal  and  vertical  lines 
need  not  actually  be  drawn;  it  is  sufficient  to  locate  only  the  points 
N  and  S.  If  the  drawing  is  done  on  squared  paper  the  horizontal 

1  A  normal  cross  section  is  a  section  made  by  a  cutting  plane  perpendicular  to  the 
axis  of  the  cylinder. 


116 


MATHEMATICS 


[§62 


and  vertical  lines  of  the  paper  will  be  of  assistance;  if  the  drawing 
is  done  on  plain  paper  the  T-square  and  triangle  should  be  used. 

Exercises 

Construct  the  curve  called  for  in  exercise  3,  §  42,  by  the  method 
of  this  section.  Construct  the  complete  curve  on  a  heavy  sheet  of 
paper,  cut  out  the  inner  portion  and  fold  to  form  a  model  of  the 
saddle. 

62.  The  Pattern  for  a  Right  Circular  Cylindrical  Surface  Cut 
by  a  Plane.  In  the  preceding  section  a  method  was  given  for  laying 
out  the  saddle  for  a  circular  ventilator.  A  method  will  now  be 


FIG.  39. 

described  for  laying  out  a  pattern  for  cutting  the  sheet  iron  which 
when  rolled  up  into  a  cylindrical  surface  will  give  the  flue  fitting 
upon  this  saddle.  The  length  of  this  pattern  will  be  equal  to  the 
circumference  of  the  cylinder. 

Draw  the  line  BQC,  Fig.  39,  equal  in  length  to  one-fourth  the 
circumference  of  the  cylinder.  This  line  corresponds  to  the  arc 
BQC,  Fig.  47,  i.e.,  if  the  sheet  of  paper  upon  which  BC  is  drawn  is 
rolled  up  into  a  cylindrical  surface  such  that  lines  perpendicular  to 
BC  become  the  elements  of  the  cylinder,  BC  becomes  the  arc 
BQC  of  Fig.  47.  From  points  of  BC,  such  as  Q,  perpendiculars 
must  be  laid  off  equal  to  QP,  Fig.  47.  From  Fig.  47,  it  is  appar- 
ent that  the  distance 

QP  =  RQ  tan  a 

where  a  is  the  angle  between  the  normal  and  inclined  cutting 
planes. 

Extend  CB  to  some  convenient  point  0.  With  0  as  center  and 
with  a  radius  equal  to  the  radius  of  the  cylinder,  describe  the  arc 


§63]  THE  CIRCULAR  FUNCTIONS  117 

BQC.  Through  0  draw  the  45°  line  OK,  and  the  line  OM  making 
the  angle  MOC  equal  to  a.  By  means  of  a  protractor  divide  the 
arc  BQC  into  a  number  of  equal  parts  (nine  for  convenience). 
Divide  the  line  BC  into  the  same  number  of  equal  parts.  From 
each  point  of  division  of  the  arc,  as  Q,  draw  a  horizontal  line  inter- 
secting OK  at  N.  Through  N  draw  a  vertical  line  intersecting 
OM  at  S.  Through  S  draw  a  horizontaliine  intersecting  at  P  the 
perpendicular  erected  at  Q  of  the  line  BC.  P  is  a  point 
upon  the  desired  curve.  For,  since  QR  =  NE  =  OE,  and  since 
PQ  =  SE  =  OE  tan  a, 

PQ  =  RQ  tan  a. 

Any  number  of  points  may  be  located  in  this  way  and  a  smooth 
curve  drawn  through  them.  The  pattern  thus  obtained  may  be 
used  four  times,  giving  the  complete  pattern. 

Exercises 

1.  Construct  a  pattern  by  the  method  explained  above  using  the 
dimensions  in  the  exercise  of  the  preceding  section. 

HINT:  This  drawing  will  require  a  sheet  of  paper  larger  than 
8£  X  11  inches. 

2.  An  elbow  is  formed  by  joining  two  right  circular  cylinders  of 
equal  radii  in  such  a  way  that  their  axes  intersect  at  right  angles.     Lay 
out  a  pattern  for  the  surfaces. 

63.  Pattern  for  a  Conical  Roof.  A  pattern  is  to  be  drawn  for  a 
conical  roof,  Fig.  40.  It  is  apparent  that  the  pattern  may  be 
made  by  cutting  out  a  cir- 
cular sector  from  a  circular 
sheet  of  paper  of  radius  r. 

If  R  is  given, 

r  =  R  sec  a. 

A  > 

The  circumference  of  the 

circular  sheet  of  paper  from  FIG.  40. 

which  the  circular  sector  is 

cut,  Fig.  41,  is  then  2irR  sec  a.     The  circumference  of  the  lower 
edge  of  the  roof  is  2irR.    Hence  the  length  of  the  arc 
DE  =  2irR  sec  a  -  2-rrR 
=  2irR(sec  a  -  1). 


118  MATHEMATICS 

Then  the  angle  BOD  expressed  in  radians  is 
_27rR(sec  a  -  1) 

r 
27rfi! (sec  a  —  1) 

R  sec  a 
=  2ir(l  —  cos  a). 


Exercises 

1.  Construct  a  pattern  for  a  conical  roof  if  a,  Fig.  40,  is  30°,  and  if 
R  is  20  percent  greater  than  the  radius  used  in  exercise  1  of  the 
preceding  section. 

2.  Build  up  suitable  formulas  for  a  pattern  for  the  convex  surface 
of  a  tank  in  the  form  of  a  frustum  of  a  right  circular  cone,  if  the  radius 
of  the  larger  base  is  R,  of  the  smaller  base  is  r,  and  if  the  depth  is  H , 
Construct  a  pattern  (reduced)  if  R  =  5,  r  =  4,  and  H  =  3. 

Miscellaneous  Exercises 

1.  The  roof  rises  from  the  adjacent  sides  of  a  rectangular  house  at 
an  angle  of  30°.  Find  the  angle  which  the  corner  of  the  roof  makes 
with  the  horizontal.  Find  the  length  of  this  edge  measured  from  the 
plate  to  the  ridge,  if  the  building  is  24  X  32  feet.  What  is  the  length 
of  the  ridge  of  the  roof? 

2.  The  height  of  a  butte  is  desired.     At  a  point  B,  the  angle  of 


§63]  THE  CIRCULAR  FUNCTIONS  119 

elevation  of  the  base  of  a  tree,  A,  standing  on  the  top  of  the  butte 
was  measured  and  found  to  be  63°  18'. 7.  At  a  second  point,  C, 
in  the  same  vertical  plane  with  AB,  'at  the  same  elevation  with,  and 
200  feet  from,  the  point  B,  the  angle  of  elevation  of  A  was  found  to  be 
32°  7'. 8.  Build  up  a  formula  giving  x,  the  elevation  of  A  above 
B;  and  y,  the  horizontal  distance  of  A  from  B. 

3.  Same  as  exercise  2,  excepting'  that  instead  of  ABC  being  in  the 
same  vertical  plane  the  angle  between  the  vertical  planes  through  A 
and  B,  and  B  and  C  is  163°  5'.6. 

4.  Same  as  exercise  2,  excepting  that  the  point  C  is  10  feet  higher 
than  the  point  B. 

5.  Same  as  problem  3,  excepting  that  instead  of  B  and  C  being 
on  a  level,  the  point  C  is  10  feet  lower  than  the  point  B. 

6.  A  100-foot  tape    hangs  down  the  face  of  a  vertical  cliff.     The 
zero  mark  is  at  A,  and  the  100  mark  at  B.     At  a  point  C  in  the 
plane  below  the  cliff,  the  angle  of  elevation  of  A  is  43°  30'. 7,  and 
of  B  is  36°  20'. 7.     Find  the  elevation  of  the  point  A  above  C. 

7.  A  and  B  are  two  points  on  the  face  of  a  vertical  cliff.     At  a 
point  C  the  vertical  line  AB  subtends  an  angle  of  7°  10';  i.e.,  the  angle 
BCA  =  7°  10'.     At  a  point   D  it  subtends  an  angle  of  6°  7'.    If 
A,  B,  C  and  D  are  in  the  same  vertical  plane  and  if  C  and  D  are 
200  feet  apart  at  the  same  elevation,  find  the  elevation  of  A  above  C, 
and  the  horizontal  distance  of  A  from  C. 

8.  Same  as  exercise  7,  but  let  D  be  5  feet  lower  than  C. 

9.  Same  as  exercise  7,  but  let  the  vertical  plane  through  CD  make 
an  angle  of  137°  10'. 7  with  the  vertical  plane  through  AC. 

10.  Same  as  exercise  9,  but  let  D  be  6  feet  lower  than  C. 

11.  A,  B,  C,  D,  E,  and  F  are  the  vertices  of  a  six-sided  field.     Find 
the  area  in  acres  if: 

AB  =  162.75  rods. 
BC  =  96.63  rods. 
CD  =  86.73  rods. 
DE  =  100.75  rods. 
EF  =  101.62  rods. 
FA  =  98.76  rods. 
AC  =  200.60  rods. 
AD  =  156.73  rods. 
AE  =  200.00  rods. 

12.  The  line  AB  runs  north  and  south.     The  line  AC  makes  an 
angle  of  52°  8'. 6  with  AB.    Locate  the  line  BC  perpendicular  to 
AB  30  that  the  area  ABC  shall  be  1  acre. 


120  MATHEMATICS  [§63 

13.  Same  as  exercise  12,   excepting  that  the  angle  ABC  is  to  be 
29°  17'.  6  instead  of  90°. 

14.  The  area,  in  acres,  of  a  four-sided  plot  of  land  with  vertices 
M,  N,  0,  and  P,  is  desired.     The  side  MP  crosses  a  low  swamp  so 
that  it  can  not  be   chained.     The   sides   MN,   NO,   and  OP  are 
measured  and  found  to  be,  respectively,  798.8  feet,  7103.7  feet  and 
2181.1  feet.     The  angles  PMN  and  NOP  are  measured  and  found 
to  be,  respectively,  76°  47'.6,  and  126°  10'.2. 

HINT:    Draw  the  diagonal  NP  and  find  the  area  of  each  triangle. 

15.  Same  as  exercise  14,  but  instead  of  measuring  the  angle  PMN  the 
angle  MNP  was  measured,  and  found  to  be  63°  42'.6. 

HINT:     Compute  the  length  of  NP  as  a  side  of  the  triangle  NOP. 

16.  Same  as  exercise  14,  but  angle  ONM  was  measured  instead  of 
angle,  PMN  and  found  to  be  127°  0'.3. 

HINT:    Compute  the  size  of  the  angle  ONP  as  an  angle  of  the 
triangle  NOP. 

17.  M,   N,   0,   and   P    are    the    vertices   of    a   four-sided    field. 
NO  =  3172.6      feet,      Z  MNP  =  43°    21'.3,      ZPNO  =  51°     2'.1, 
Z  NOM  =  55°  6'.5,  and  Z  MOP  =  47°  52'.6.     Find  the  area  of  the 
field  in  acres. 

HINT:     Calculate  MN  and  angle  MNO.     Find  the   area  of  the 
triangles  NOP  and  MNP. 

18.  Find  the  volume,  in  cubic  yards,  of  a  right  circular  cylindrical 
shell  of  uniform  thickness.     To  make  the  problem  specific,  let  us  find 
the  number  of  yards  of  concrete  in  the  vertical  wall  of  a  circular  con- 
crete silo,  assuming,  first,  that  there  are  no  openings.     Let  R  be  the 
outer   radius,  d   the   thickness  of   the   wall   and    H  the  height,   all 
measured  in  feet.     From  the  horizontal  cross  section  of  the  silo,  it 
is  seen  that  the  area  of  the  cross  section  of  the  wall  is  the  difference  of 
the  areas  of  the  two  circles,  one  with  radius  R  and  the  other  with  ra- 
dius R  -  d.     This  area  A  =  irRz  -  ir(R  -  d)2  =  TrlR2  -  (R  -  d)2]. 
The  quantity  within  the  parentheses  is  the  difference  between  two 
squares,  and  may  be  resolved  into  the  product  of  two  factors,  the  one 
the    sum    and    the    other    the    difference    of    the    numbers.     Then 
A  =  ir(2R  —  d~)(d).     This  area  multiplied  by  the  height  and  divided 
by  27  gives  the  volume,  V,  of  the  wall  in  cubic  yards.     Then 

ir(2R  - 


27 

a  formula  suitable  for  logarithmic  computation. 

ILLUSTRATION  :     Suppose  the  outside  diameter  of  a  silo  is  13  feet, 


§63]  THE  CIRCULAR  FUNCTIONS  121 

the  thickness  of  the  wall  7  inches  (0.5833  feet,  from  Table  XVII),  and 
heighl;  35  feet.     Then 

H  =  35 
d  =  0.5833 
2R  -  d  =  12.4167 
log  H  =  1.5441 
log  d  =  9.7659  -  10 
log  (2R  -  d)  =  1.0940 
log  TT  =  0.4971 
log  27  =  1.4314 

log  V  =  1.4697 

V  =  29.49  cubic  yards. 

If  only  a  single  computation  were  to  be  made,  the  formula  for 
volume  could  be  built  up  using  the  data  directly.  For  the  above 
measurements : 

7r(6.5)2  =  area  of  larger  circle 
7r(6.5  —  0.5833)2  =  area  of  inner  circle 
7r(6.52  -  5.91672)  =  7r(6.5  +  5.9167)  (6.5  -  5.9167) 
=  7r(12.4167)  (0.5833) 
=  area  of  cross  section 
Then 

V  =  35rr (12.4167)  (0.5833)  -h  27 
=  volume  in  cubic  yards. 

If  several  sets  of  measurements  were  given,  and  if  volumes  cor- 
responding to  all  were  to  be  computed,  it  would  be  desirable  to  derive 
first  the  general  literal  formula  and  then  substitute  the  given  nu- 
merical values.  The  example  explained  above  illustrates  how  much 
easier  it  is  to  effect  the  transformations  when  letters  rather  than 
numerical  values  are  used. 

The  student  is  reminded  that  in  general  it  is  advisable  to  represent 
the  known  and  unknown  numbers  by  letters  and  build  up,  if  possible, 
a  formula  for  each  unknown  in  terms  of  known  numbers  before  mak- 
ing numerical  substitutions. 

19.  Find  the  volume  in  cubic  yards  of  a  right  circular  cylindrical 
shell,  if  the  outer  radius  is  7.25  feet,  the  thickness  8  inches,  and  the 
length  28.5  feet. 

20.  In  exercise  18,  the  volume  of  the  opening  must  be  deducte 
from  the  total  volume  of  the  wall.    Let  the  sides  of  the  opening  be 
two  vertical  planes  passing,  if   extended,  through  the  axis  of  the 


122 


MATHEMATICS 


[§63 


cylinder.1  Fig.  42  represents  a  cross  section  of  the  opening.  Let  S 
be  the  width  of  the  opening  on  the  outside,  and  s  the  width  on  the 
inside.  These  distances  are  to  be  measured  along  the  arcs  of  circles 
and  not  along  the  chords. 

%RS  =  area  of  circular  sector  ABO. 
%(R  —  d)s  =  area  of  circular  sector  CDO. 
%[RS  -  (R  -  d)s]  =  area  ACDB. 


Before  the  last  equation  can  be  used  we  must  find  an  expression  for  s, 
for  this  is,  as  yet,  unknown.     From  geometry, 

s       R  -  d 
=  ? 

S         R 

or 

_  S(R  -  d) 

R  ~~' 

Substituting  for  s  in  the  above  formula  for  the  area  of  ACDB  we 
have 

K<»ACDB-i\SR   _^I^~| 

R        J 


=  ~<2J2 

ZK 

1  Thii  is  not  the  true  form  of  the  opening  in  the  silo;  the  exercise  is  only    in- 
iroduced  here  to  illustrate  how  the  formula  is  built  up. 


THE  CIRCULAR  FUNCTIONS 


123 


Multiplying  this  area  by  the  height,  h,  of  the  opening,  and  dividing 
by  27,  the  volume,  v,  in  cubic  yards  is 


54  R 

ILLUSTRATION:     Find  v,  if  S  =  38  inches  (3.1667  feet)  and  h  =  29.5 
feet,  for  the  cylinder  whose  volume  was  computed  in  exercise  18. 

logd  =  9.7659  -  10 

log  (2R  -  d)  =  1.0940 

logS  =  0.5006 

log  h  =  1.4698 

log  numerator  =  2.8303 

log  R  =  0.8129 

log  54  =  1.7324 

log  denominator  =  2  .  5453 

log  v  =  0.2850 

v  =  1  .  928  cubic  yards. 

Find    v    for    the    set    of    dimensions    given    in    exercise    19   if 
S  =  40  inches,  and  h  =  21  feet. 

21.  Let  Fig.  43  represent  a  vertical  cross  section  of  a  conical  con- 
crete roof  of  a  silo.  Find  the  number  of  cubic  yards  of  concrete  in 
the  roof,  assuming  no  open- 
ing. Let  k,  r,  and  a  be  the 
known  dimensions  as  repre- 
sented in  the  figure.  The 
volume  sought  is  the  differ- 
ence  between  the  volume 
of  AA'C'B'BDA  and  of 
ACBDA.  The  first-named 
volume,  however,  is  the  sum 
of  the  volume  of  the  right 
circular  cone  A'C'B'D'A' 

and  the  volume  of  the  circular  disc  AA'D'B'BDA;  and  since 
the  volume  of  A'C'B'D'A'  is  equal  to  the  volume  of  the  right  cir- 
cular cone  ACBDA,  the  required  volume  is  equal  to  the  volume  of 

the  circular  disc  AA'D'B'BDA,  or  *r*(BB').  But,  BB'  =  ^-^- 
The  final  formula  for  the  volume,  u,  then  becomes 


u  = 


FIG.  43. 


COS  a 


Find  u  if  r  =6.25  feet,  k  =  3  inches,  and  a  =  30°. 


124 


MATHEMATICS 


22.  Fig.  44  represents  a  water  tank  in  the  form  of  a  frustum  of  a 
right  circular  cone.  Let  R  and  r  be,  respectively,  the  radii  of  the 
lower  and  upper  bases.  Let  H  be  the  altitude.  Let  x  be  the  depth 
of  the  water  in  the  tank.  AB  is  a  vertical  glass  tube  connected  with 
the  tank  as  shown  in  the  figure.  The  surface  of  the  water  in  the 
tube  at  C  rises  to  a  level  with  the  surf  ace  of  the  water  in  the  tank; 
and  as  water  is  added  to  or  taken  from  the  tank,  C  will  rise  or  fall. 


FIG.  44. 


If  a  uniform  scale  with  its  zero  on  a  level  with  the  bottom  of  the 
tank  be  placed  behind  the  tube,  the  reading  upon  it  as  indicated  by  C 
will  give  the  depth  of  the  water  within  the  tank,  and  will  not  give, 
excepting  indirectly,  the  volume  of  water  within  the  tank. 

The  problem  is  to  construct  a  non-uniform  scale,  such  that,  when 
placed  behind  the  tube,  AB,  it  shall  give  the  direct  reading  for  the 
volume  of  water  in  the  tank.  It  is  at  once  evident  that  the  zero  for 
the  scale  must  be  on  a  level  with  the  bottom  of  the  tank,  and  that 
the  divisions  corresponding  to  equal  increments  of  volume  of  water 
are  farther  and  farther  apart  as  the  upper  end  of  the  scale  is  ap- 
proached. This  scale  may  be  constructed  by  running  known  quan- 
tities of  water  into  the  tank  and  marking  the  position  of  the  point  C; 
or  it  may  be  constructed  by  computing,  from  a  formula,  values  of  x 


THE  CIRCULAR  FUNCTIONS  125 

for  given  values  of  the  volume.     Let  V  represent  the  volume  of  water 
in  the  tank.     Show  that 

ir7?3       Trjri  rJ?         ~\3 
V  = 


_  «*f  I"B  _    I 

~   3    Ik   ~X\ 


3fc 
•where  k  =  (R  -  r)/H. 

Before    solving    for    x,    make    the  following   abbreviations.     Let 

irR3  7T/C2         1  R 

—  =  a,   -  —  =  — ,  and  —  =  y.     Then 


or 


x  =  y  -       /3(a  -  F), 


where  a,  /3,  and  7  are  constants  for  a  given  tank.  From  this  formula 
we  may  compute  x  corresponding  to  any  value  or  set  of  values  for  V. 
These  values  of  x  are  to  be  measured  off  from  the  zero  upon  the  scale, 
and  the  points  thus  located  are  to  be  marked,  not  with  the  x  values, 
but  with  the  values  given  to  V. 

Compute  x  for  every  interval  of  20  cubic  feet  in  the  value  of  V 
ranging  from  0  to  600  cubic  feet,  if  R  =  5  feet,  r  =  3  feet,  and  H  =  10 
feet.  With  the  values  of  x  computed  construct  a  miniature  scale  on 
which  1  inch  represents  1  foot.  Mark  the  divisions  representing  the 
even  hundreds  of  cubic  feet  much  longer  than  the  subdivisions  rep- 
resenting the  even  twenties.  Number  only  the  longer  divisions,  as 
0,  100,  200,  300,  .  .  . 

In  performing  the  numerical  work,  follow  the  outline  given  below. 
Place  the  values  of  V,  as  20,  40,  60,  80,  100,  120,  140,  .  .  .  ,  600  upon 
the  first  horizontal  line  of  a  sheet  of  paper  form  M7.  Subtract  these 
values  from  the  value  of  a,  and  place  the  differences  on  the  second  hori- 
zontal line.  The  values  of  a,  y,  and  log  /3  should  be  computed  at 
some  out-of-the-way  place  upon  the  lower  part  of  the  calculation 
sheet.  In  subtracting  V  from  a,  write  a  at  the  lower  edge  of  a  slip 
of  paper  and  by  sliding  the  slip  along,  the  thirty  subtractions  may 
be  performed  with  the  value  of  a  written  but  once.  Upon  the  third 
horizontal  line  place  the  logarithms  of  a  —  V,  to  which  add  the 
logarithm  of  /3  and  place  the  sums  upon  the  fourth  line.  Divide  these 
logarithms  by  3,  placing  the  quotients  upon  the  fifth  line.  Take  out 
the  numbers  corresponding  to  the  logarithms  given  on  line  5,  writing 
them  on  line  6.  Subtract  these  numbers  from  y,  placing  the  differ- 
ences on  the  seventh  line,  and  the  numerical  work  is  complete. 

In  performing  the  numerical  work,  several  sheets  of  computation 


126  MATHEMATICS  [§63 

paper  may  be  used,  or  by  dropping  down  about  ten  lines,  one  sheet 
may  be  crossed  several  times  with  values  of  V. 

Before  constructing  the  miniature  scale,  it  is  advisable  first  to  plot 
the  values  for  V  and  x  upon  a  sheet  of  squared  paper.  By  so  doing 
any  appreciable  error  in  an  x  will  be  apparent  by  the  corresponding 
point  not  falling  upon  a  smooth  curve  drawn  through  the  plotted 
points.  The  value  of  x  for  any  point  not  falling  upon  the  curve  should 
be  recomputed. 

The  values  for  a,  ft,  and  log  7  should  be  computed  several  times 
to  insure  their  correctness. 


1000ft. 

FIG.  45. 

23.  To  find  the  distance  from  the  flagstaff  on  University  Hall  to 
the  flagstaff  on  the  Capitol  building,  a  base  line,  AB,  1000  feet  long, 
Fig.  45,  was  laid  off  on  Lakeside  Ave.,  South  Madison,  and  the 
following  angles  measured: 

CBA  =  109°  53'  15". 

DBA  =    67°  35'  45". 

DAB  =  105°    1'  30". 

CAB  =    63°    2'  45". 


§63] 


THE  CIRCULAR  FUNCTIONS 


127 


From  the  triangle  ADB  compute  the  length  of  AD  and  of  DB. 
From  the  triangle  ACB  compute  the  length  of  AC  and  of  CB. 
Compute  x  from  the  triangle  ACD,  and  also  from  the  triangle  DBC, 
and  take  the  mean  of  the  two  values. 

24.  To  find  the  height  of  the  ball  on  the  flagstaff  on  University 
Hall  above  the  water  table  of  North  Hall,  the  following  measurements 
were  taken  with  the  transit  placed  at  the  points  A  and  B,  100  feet 
apart,  Fig.  46. 


Si  Ball 


100  ft. 

FIG.  46. 


The  angle  of   elevation   of  S  at  A,     Z  SAS"  =  27°  51'  30". 

The  angle   of  elevation   of   S  at  B,   Z  SBS'  =  28°  35'  30". 

The  horizontal  angle  S"AB  =  75°  31'. 

The  horizontal  angle  S'BA  =  84°  21'. 

At  A  the  transit  was  3.33  feet,  and  at  B,  2.71  feet  above  the  water 
table  of  North  Hall. 

Using  the  law  of  sines,  calculate  the  logarithm  of  the  length  BS', 
and  of  AS".  From  the  triangle  BS'S  calculate  the  length  of  SS', 
and  from  the  triangle  AS"S  the  length  of  SS".  To  SS'  add  2.71, 
and  to  SS"  add  3.33.  Take  the  mean  of  the  last  two  sums  as  the 
height  of  S  above  the  water  table. 


CHAPTER  IV 
[THE  ELLIPSE] 

64.  The  Ellipse  Defined.  We  shall  define  the  ellipse  as  the 
curve  of  intersection  of  a  plane  with  the  surface  of  a  right  circular 
cylinder.  This  definition  includes  the  circle  as  a  special  case  of 
an  ellipse. 


FIG.  47. 

In  laying  out  the  pattern  for  the  saddle  of  a  circular  ventilator, 
exercise  3,  §  42,  an  ellipse  was  constructed  by  drawing  a  curve 
through  plotted  points.  In  this  exercise  the  equation  of  the  curve 
was  derived  for  the  case  in  which  the  cutting  plane,  the  roof,  made 
an  angle  of  60°  with  the  axis  of  the  cylinder.  The  method  of  find- 
ing the  equation  is,  however,  the  same  for  all  values  of  the  angle. 

128 


§65]  THE  ELLIPSE  129 

65.  The  Equation  of  the  Ellipse.  Let  BCB'DB,  Fig.  47,  be  a 
normal  cross  section  of  a  right  circular  cylinder.  The  curve 
BCB'DB  is  then  a  circle.  Let  BAB'A'B  be  the  intersection  of 
the  cylinder  with  a  plane  inclined  to  the  normal  cutting  plane  at 
an  angle  a.  The  curve  BAB'A'B  is  then,  by  definition,  an  ellipse. 
Let  the  normal  and  the  inclined  cutting  planes  intersect  in  the 
diameter  of  the  cylinder,  BB'.  Let  the  radius  of  the  cylinder 
be  6. 

In  the  plane  AB'A'BA  let  0  be  the  origin,  OB  the  positive  direc- 
tion of  the  X-axis,  and  OA  the  positive  direction  of  the  7-axis, 
of  a  rectangular  coordinate  system.  If  P  be  any  point  on  the 
curve  AB'A'BA,  OR  is  its  z-coordinate  and  RP  its  ^/-coordinate. 

Let  Q  be  the  point  of  intersection  of  the  normal  section  with  the 
element  passing  through  P. 

Then  OR2  +  RQ2  =  OQ2, 

or 

X2  +   RQ2    =    b2, 

or,  since 

RQ  =  RP  cos  PRQ  =  y  cos  a, 

x2  +  y2  cos2  a  =  b2, 
or 

x2       y2  cos2  a  _ 
b2         "b2         ~~    ' 

which  is  the  equation  of  the  ellipse. 

When  a  is  zero,  this  equation  reduces  to  z2  +  y2  =  b2,  the 
equation  of  a  circle,  the  normal  cross  section  of  the  cylinder.  As 
a  increases  from  0°  to  90°,  cos  a  decreases  from  1  to  0,  and  the 

cos2  1 

coefficient  of  y2,     ,2    >  decreases  from  ?2  to  0.     When    a  =  90°, 

the  equation  of  the  ellipse  reduces  to  v^  =  1,  or    x2  =  b2,   or 

z2  —  62  =  0,  or  (x  —  b)  (x  +  b)  =  0.  The  last  equation  is  equiva- 
lent to  the  two  equations,  z  —  6  =  0,  and  z  +  6  =  0,  or  x  =  b 
and  z  =  —  b;  for  if  the  product  of  two  expressions  is  equal  to 

9 


130  MATHEMATICS  [§66 

zero,  the  equation  is  satisfied  by  equating  either  factor  to  zero.1 
Each  equation,  x  =  b  and  x  =  —  b,  represents  a  straight  line 
parallel  to  the  F-axis,  6  units  from  it,  but  upon  opposite  sides  of 
the  origin.  This  is  exactly  what  sliould  be  expected,  for  when 
a  =  90°,  the  cutting  plane  becomes  parallel  to  the  elements  of  the 
cylinder  and  the  curve  of  intersection  reduces  to  two  straight 
lines. 

66.  General  Shape  of  the  Ellipse.    Let  OA,  Fig.  47,  be  repre- 
sented by  a.    Since  OC  =  6,  and  since  /  AOC  =  a,  it  follows 

that  -  =  cos  a  or  -     -  =  a.     Substituting    this    value    for 
a  cos  a 

in  the  equation  of  the  ellipse.,  it  reduces2  to 


cos  a 


or,  upon  solving  for  y, 

y  =  ±1  V&2  -  x2 

This  equation  shows  that  for  every  value  assigned  to  x  there  cor- 
respond two  values  for  y,  numerically  equal  but  with  opposite 

1  It  will  be  recalled  that  the  method  of  replacing  one  equation  by  two  equations 
is  that  which  may  be  used  in  solving  quadratic  equations.  (See  §  15). 

Let  x2  —  x  —  6  =  0  be  the  equation  to  be  solved.  By  factoring  the  left-hand 
side,  this  equation  may  be  written  (x  +  2)(z  —  3)  =  0.  To  solve  an  equation  for 
x  is  to  find  all  values  such  that  when  they  are  substituted  for  x  in  the  equation,  the 
equation  is  satisfied,  i.e.,  the  left-hand  side  reduces  identically  equal  to  the 
right-hand  side.  To  solve  then  the  above  equation,  considering  it  written  in  the 
second  form,  means  to  find  all  values  which  when  substituted  for  x  reduce  the  left- 
hand  side  to  zero.  To  make  the  left-hand  member  of  this  equation  zero  in  all 
possible  ways  is  to  equate  each  factor  to  zero.  Thus  x  +  2  =  0  and  x  —  3  =  0. 
But,  for  x  +  2  to  be  equal  to  zero,  x  must  equal  —  2,  and  for  x  —  3  to  be  equal  to 
zero,  x  must  equal  3.  Therefore,  —  2  and  3  are  the  only  values  of  x  which  when 
substituted  for  x  in  the  original  equation  satisfy  that  equation,  i.e.,  —  2  and  3  is 
the  solution  of  the  equation. 

In  general,  if  the  right-hand  member  of  an  equation  is  zero,  and  if  the  left-hand 
member  breaks  up  into  two  or  more  factors,  the  equation  is  equivalent  to  the  system 
of  equations  formed  by  equating  each  factor  to  zero.  Finding  all  solutions  of  the 
system  of  equations  gives  all  solutions  of  the  original  equation. 

1  It  must  be  remembered  that  Fig.  47  is  only  a  perspective  drawing  of  the 
cylinder  and  the  two  cuttng  planes,  and  that  we  are  at  present  concerned  with  the 
curve  AB'A'BA  as  it  appears  upon  the  plane  of  the  paper.  We  must  imagine  that 
we  are  looking  down  upon  the  inclined  cutting  plane  in  such  a  way  that  OB  extends 
to  our  right  and  that  OA  is  the  upward  vertical  direction.  The  outline  would  then 
appear  to  us  as  in  Fig.  48. 


§66]  THE  ELLIPSE  131 

signs.  Geometrically  this  means  that  the  curve  is  symmetrical 
with  respect  to  the  axis  of  x.  For  if  the  curve  were  plotted  point 
by  point,  the  two  numerically  equal  values  of  y,  corresponding  to 
a  single  value  of  x,  show  that  there  are  two  points  upon  the  curve 
at  the  same  distance  from  the  .XT-axis;  and  the  opposite  signs  of 
the  two  values  of  y  show  that  one  point  is  above  the  X-axis  while 
the  other  is  below.  These  two  points,  since  they  correspond  to  a 
single  value  of  x,  are  located  upon  the  same  side  of,  and  at  the 
same  distance  from,  the  F-axis.  The  line  joining  the  two  points 
is  then  perpendicular  to,  and  bisected  by,  the  X-axis.  Thus  the 
two  points  are  symmetrical  with  respect  to  that  axis.  Since  all 
points  of  the  curve  occur  in  pairs  symmetrically  situated  with 
respect  to  the  X-axis,  the  curve  is  said  to  be  symmetrical  with 
respect  to  that  axis.  As  an  illustration:  let  the  radius,  6,  of  the 
cylinder  be  2  feet,  let  the  angle  between  the  normal  and  the  in- 
clined cutting  planes,  a,  be  30.° 

Then 

__  _ 

"  "       ~°  ~         ~          '  a 


.—  ,— 

=  ±      V62  —  x2  =  ±  —  --  V  4  —  x2. 


When  x  =  1  the  last  equation  gives 
.  2V3    ,— 


Thus,  (1,  2)  and  (1,  —  2)  are  two  points  on  the  curve,  each,  one 
unit  to  the  right  of  the  7-axis.  One  point  is  two  units  above  the 
X-axis,  and  the  other  is  the  same  distance  below. 

From  the  equation  y  =  ±  -r  V&2  —  a;2  it  is  seen  that  y  increases 

in  numerical  value  as  the  quantity  under  the  radical  increases,  and 
decreases  as  the  quantity  under  the  radical  decreases.  The 
quantity  under  the  radical,  &2  —  xz,  increases  as  x  decreases  in 
numerical  value,  and  decreases  as  x  increases.  The  value  of 
62  —  x2,  which  is  62  when  x  is  zero,  decreases  to  zero  as  x  increases 
to  6,  and  becomes  negative  when  the  value  of  x  exceeds  that  of  b. 


132 


MATHEMATICS 


[§66 


The  numerical  value  of  y,  which  has  a  maximum  when  x  equals 
zero,  decreases  to  zero  as  x  increases  to  b.  When  the  value  of  x 
exceeds  b,  y  becomes  imaginary,1  since  the  number  under  the  radical 
sign  is  then  negative.  This  says,  geometrically  that  for  all  values 
of  x  greater  than  6,  there  is  no  real  value  of  y  and  no  point  can 
s  A  T  be  found  whose  coordi- 

nates satisfy  the  given 
equation;  or  the  curve 
does  not  extend  to  the  right 
of  a  vertical  line,  b  units  to 
the  right  of  the  F-axis. 

The  curve  is  symmetrical 
with  respect  to  the  F-axis. 
This  fact  is  shown  by  its 
equation,  since  x  occurs 
only  as  the  square.  If 
two  values  are  assigned  to 
x,  numerically  equal  but 
with  opposite  signs,  the 
corresponding  pairs  of 
values  for  y  are  the  same. 
Which  says,  geometrically 
that  for  any  point  of  the 
curve  there  is  another 
point  symmetrical  to  the 
first  point  with  respect  to 
the  F-axis.2 

From  the  preceding  discussion  it  is  seen  that  the  curve  does 
not  extend  to  the  right  of  the  line  T'T,  Fig.  48,  nor  to  the  left 
of  the  line  SS';  that  the  curve  does  not  extend  above  the  line  ST 
nor  below  the  line  S'T'.  Therefore  the  curve  is  entirely  within 
the  rectangle  of  dimensions  2a  and  26.  These  conclusions  are 
geometrically  apparent  from  Fig.  47. 

1  The  even  root,  as  the  square  root,  the  fourth  root,  the  sixth  root,  etc.,  of  a  nega- 
tive quantity  is  called  imaginary. 

2  That  the  curve  is  symmetrical  with  respect  to  the  F-axis  may  also  be  shown  by 
first  solving  the  equation  for  x.     For,  in  this  form,  it  is  readily  seen  that,  for  any 
value  of  y,  numerically  less  than  a,  these  are  two  numerically  equal  values  of  x,  one 
positive  and  one  negative. 


S' 


§67] 


THE  ELLIPSE 


133 


OE,(  =  b)  and  OA,(  =  a)  are  called  the  semi-axes  of  the  ellipse.1 

67.  A  Geometric  Method  of  Locating  Points  upon  an  Ellipse. 

With  0  as  center,  Fig.  49,  and  with  OB,  or  b,  and  OA,  or  a,  as 

radii,  draw  two  circles.     From  0  draw  any  line  making  an  angle  6 

with  the  positive  direction  of  the  axis  of  x,  cutting  the  smaller 


FIG.  49. 

circle  at  the  point  M  and  the  larger  circle  at  the  point  N.  Through 
M  draw  a  line  parallel  to  the  F-axis,  and  through  N  a  line  parallel 
to  the  .X"-axis.  We  shall  show  that  P,  their  point  of  intersection, 
is  a  point  upon  the  ellipse.  The  x-coordinate,  OS,  of  the  point  P 

1  In  text-books  on  Analytic  Geometry  it  is  customary  to  take  the  major  semi-axis 
of  the  ellipse  coinciding  with  the  axis  of  x.  The  ellipse  would  then  appear  as  Fig. 
48  rotated  through  90°.  Its  equation  is 


where  6  >  a. 


134  MATHEMATICS  [§68 

is  OM  cos  6,  or  x  =  b  cos  6.    The  T/- coordinate,  SP,  or  TN,  of 
the  point  P  is  ON  sin  6,ory  =  a  sin  0.     From  these  two  equations , 


-T  =  cos  6 

0 


and 

V 


=  sin 
a 


+    i  -  cos2  0  +  sin2 


J 


or,  since  sin2  0  +  cos2  0=1, 

This  is  the  equation  of  the  ellipse.  Therefore,  P  is  a  point  upon 
an  ellipse.  By  the  above  method  any  number  of  points  may  be  lo- 
cated upon  the  ellipse  and  the  curve  drawn  through  them. 

68.  The  Foci;  the  Eccentricity;  the  Sum  of  the  Focal  Radii. 
At  B,  Fig.  49,  draw  a  line  tangent  to  the  smaller  circle  cutting 
the  larger  circle  at  the  points  E  and  E'.  From  E  and  E'  drop 
perpendiculars  upon  the  F-axis  meeting  it  at  the  points  F  and 
F'.  F  and  F'  are  called  the  foci  of  the  ellipse,  and  the  lines  FP  and 
F'P,  lines  drawn  from  the  foci  to  any  point  P  of  the  ellipse,  are 
called  the  focal  radii  for  the  point  P. 

We  shall  show  that  FP  +  F'P  =  2a  (a  constant  for  all  points 
upon  the  ellipse). 

Let  FP  be  abbreviated  by  n  andP'Pby  rt,  and  OF  (  =  OF') 
by  c.  Then 

or 

ri»  =  x2  +  y2  +  c2  -  2cy, 
or 

ri2  =  b2  cos2  0  +  a2  sin2  0  +  c2  -  2ca  sin  0. 

jt  is  easily  seen  from  the  rectangle  OBEF  that  c2  =  a2  —  b2, 
whence, 

ri2  =  b2  cos2  0  +  a2  sin2  0  +  a2  -  b2  -  2ac  sin  0, 
or  since  cos2  0=1  —  sin2  0, 

ri2  =  62(1  -  sin2  0)  +  a2  sin2  0  +  a2  -  b2  -  2ac  sin  0, 


§68]  THE  ELLIPSE  135 

or  ri2  =  (a2  -  62)  sin2  0  -  2ac  sin  0  +  a2 

n2  =  a2  -  2ac  sin  0  -f  c2  sin2  0 
ri  =  a  —  c  sin  0. 

Similarly  it  can  be  shown  that 

TZ  =  a  +  c  sin  0. 
Upon  adding, 

ri  +  r2  =  2a, 
or  the  sum  of  the  lengths  of  the  two  focal  radii  is  a  constant. 

This  property  of  the  ellipse  furnishes  an  easy  method  for  its 
construction.  From  the  relation  c2  =  a2  —  62,  locate  the  foci, 
F  and  F'.  Fasten  the  ends  of  a  string,  whose  length  is  2a,  at  the 
points  F  and  f".  Place  a  pencil  point  in  the  loop  of  this  string 
and  draw  taut.  By  moving  the  pencil,  allowing  the  string  to  slip 
around  it,  an  ellipse  will  be  described. 

We  shall  prove  the  converse  of  the  above  theorem:  The  locus 
of  a  point,  moving  in  a  plane  such  that  the  sum  of  its  distances 
from  two  fixed  points  remains  constant,  is  an  ellipse. 

Let  F  and  F'  be  the  fixed  points,  and  P  any  point  upon  the 
curve.  Let  the  F-axis  pass  through  F  and  F'.  Let  0,  the  mid- 
point of  FF',  be  the  origin.  Let  the  distance  OF  (  =  OF')  be  rep- 
resented by  c.  Let  FP  be  represented  by  rt  and  F'P  by  r2. 

n2  =  xz  +  (c  -  7/)2  (1) 

r22  =  z2  +  (c  +  y)2.  (2) 

By  adding, 

n2  +  r22  =  2(z2  +  ?/2  +  c2).  (3) 

By  subtracting, 

r22  —  ri2  =  4cy.  (4) 

fi  +  r-i  =  2a  (constant  sum)  by  hypothesis.  (5) 

Dividing  (4)  by  (5), 


By  adding  (5)  and  (6), 

r2  =  a+f-  (7) 

By  subtracting  (6)  from  (5), 

*  — *  (8) 


136  MATHEMATICS  [§69 

Squaring  (7)  and  (8)  and  adding, 

ri2  +  r22  =  2(a2  +  c2?/2/a2).  (9) 

Substituting  in  the  first  member  of  (3)  and  canceling  the  common 
factor  2, 

C2W2 

a2  +  —  =  x2  +  y2  +  c2. 
Clearing  of  fractions, 

or 

a2z2  +  (a2  -  c%2  =  a2(a2  -  c2).  (10) 

Dividing  (10)  by  the  quantity  a2(a2  -  c2), 

-Y.2  /j;2 

x     i  y_  _  j 

a2  —  c2      a2 
But,  this  equation  is  of  the  form 

fc2  +  a2  =  L 

which  we  have  shown  to  be  the  equation  of  the  ellipse. 

The  distance  OF  =  OF'  =  c  is  called  the  focal  distance.  The 
ratio,  c/a,  which  is  never  greater  than  unity,  is  called  the  ec- 
centricity of  the  ellipse.  Referring  to  Fig.  47  the  focal  distance 
is  equal  to  AC,  and  the  eccentricity  is  the  sin  a.  A  circle  then 
is  an  ellipse  with  zero  eccentricity. 

69.  The  Area  of  the  Ellipse.  Let  ABCD  and  ABC'D',  Fig.  50, 
be  two  planes  inclined  to  each  other  at  an  angle  a.  Let  EFG'H' 
be  a  rectangle  in  the  plane  ABC'D'  with  its  base  EF  in  (or  parallel 
to)  the  line  of  intersection  of  the  two  planes.  Let  EFGH  be  the 
orthographic  projection  of  EFG'H'  upon  the  plane  ABCD.1  Since 
EH  =  EH'  cos  a,  the  area  of  EFGH  =  (area  of  EFG'H')  cos  a. 

If,   instead  of  a  rectangle,   an  irregular  area,   as  E'F'G'H', 

1  If  from  a  point  a  perpendicular  is  let  fall  upon  a  plane,  the  foot  of  the  perpendicular 
is  called  the  orthographic  projection  of  the  point  upon  the  plane.  If  from  every 
point  of  a  curve  a  perpendicular  is  let  fall  upon  a  plane,  the  locus  of  the  foot  of  this 
perpendicular  upon  the  plane  is  called  the  orthographic  projection  of  the  curve  upon 
the  plane.  If  the  bounding  curve  or  curves  of  any  surface  be  projected  orthographic- 
ally  upon  a  plane,  the  area  bounded  by  the  projection  is  called  the  orthographic 
projection  of  the  area  of  the  given  surface  upon  the  plane. 


§69] 


THE  ELLIPSE 


137 


Fig.  51,  on  the  plane  ABC'D'  be  projected  orthographically  upon 
the  plane  ABCD,  divide  the  projected  area  into  strips  by  equidis- 


FIG.  50. 


FIG.  51. 

tant  parallel  lines  drawn  perpendicular  to  AB,  and  from   the 
points  of  intersection  of  each  of  these  lines  with  the  boundary  of 


138  MATHEMATICS  [§70 

the  area  draw  lines  to  the  right  until  they  meet  the  next  parallel. 
In  this  way,  on  the  plane  ABC'D',  is  formed  a  series  of  rectangles 
which  project  into  another  series  of  rectangles  on  the  plane  ABCD. 
From  what  was  shown  above  it  is  seen  that  the  sum  of  the  areas 
of  the  projections  of  these  rectangles  upon  the  plane  ABCD  is 
equal  to  the  sum  of  the  areas  of  the  projected  rectangles  multi- 
plied by  the  cos  DAD'. 

As  the  distance  between  the  parallel  lines  is  made  smaller  and 
smaller,  i.e.,  as  the  number  of  strips  is  increased  indefinitely, 
the  sum  of  the  areas  of  the  rectangles  in  the  plane  ABC'D'  ap- 
proaches nearer  and  nearer  the  irregular  area  E'F'G'H' ;  and  the  sum 
of  the  areas  of  the  rectangles  in  the  plane  ABCD  approaches  nearer 
and  nearer  the  area  EFGH,  the  proj  ection  of  E'F'G'H' .  Therefore, 
area  EFGH  =  (area  E'F'G'H')  cos  DAD'.  Then  (Fig.  47)  the 
area  of  the  circle  BCB'D  =  (area  of  the  ellipse  ABDB')  cos  a,  or, 
since  cos  a  =  b/a,  and  since  area  of  the  circle  =  irb2, 

•jrb2  =  (area  of  ellipse)—' 

or 

area  of  ellipse  =  Trba 

70.  The  Orthographic  Projection  of  a  Circle.  Let  ABCD, 
Fig.  52a,  be  a  circle  with  radius  a.  Let  AOC,  Fig.  526,  be  an 
end  elevation  of  the  same  circle.  Rotate  this  circle  about  BOD 
as  an  axis  through  an  angle,  /3,  to  the  position  A"OC".  Project 
the  rotated  circle  upon  its  original  plane,  into  the  curve  A'BC'D. 
We  shall  show  that  A'BC'D  is  an  ellipse.  Take  any  point  P  upon 
the  original  circle.  It  rotates  into  the  point  P",  and  P"  projects 
intoP'.  The  equation  of  the  circle  is  x*  +  y2  =  a2,  where  x  =  MP. 
To  get  the  equation  for  the  curve  A'BC'D  replace  MP  by  its 
equal  MP'  /cos  0.  (See  Fig.  526.)  Whence, 

MP'2 

-— 4-  v2  =  a2 

cos2 ,3  +y 

Since  MP'  is  the  x-coordinate  of  P', 


§71] 
or 


THE  ELLIPSE 


139 


a2  cos2  /3  ""  a2 
Replacing  a  cos  /3  by  &(  =  OC"), 

a;2      v2 


the  equation  of  an  ellipse.    The  focal  distance,  Va2  —  ft2,  is  equal 
to  C'C",  and  the  eccentricity  c/a  =  sin  ft.1 


FIG.  52. 

[71.]  Tangent  to  an  Ellipse.  Let  PTQ,  Fig.  47,  be  a  tangent 
plane  to  the  cylinder  with  PQ  the  line  of  tangency.  Let  QT  and 
PT  be,  respectively,  the  lines  of  intersection  of  the  tangent  plane 
with  the  horizontal  and  the  inclined  cutting  planes.  TQ  is  tangent 
to  the  circle  DBQCB'  at  the  point  Q,  and  TP  is  tangent  to  the  el- 
lipse BPAB'A'  at  the  point  P.  These  tangent  lines  meet  at  the 
point  T  upon  the  shorter  axis  of  the  ellipse  produced.  This  fact 
will  be  made  use  of  in  drawing  the  tangent  to  an  ellipse. 

Let  ABA'B',  Fig.  53,  be  an  ellipse  to  which  a  tangent  line  is  to 
be  drawn  at  the  point  P.  Draw  the  circle  with  B'B  as  diameter. 
From  P  drop  a  perpendicular  upon  B'B  cutting  the  circle  at  Q. 

1  It  can  be  shown  that  the  section  of  a  right  circular  cone  made  by  a  plane  cutting 
all  of  its  elements  on  the  same  side  of  the  vertex  is  an  ellipse. 


140 


MATHEMATICS 


[§71 


Through  Q  draw  QT  tangent  to  the  circle  (perpendicular  to  OQ). 
Draw  TP,  which  will  be  tangent  to  the  ellipse  at  the  point  P. 


FIG.  53. 


FIG.  54. 

It  is  left  for  the  student  to  show  that  the  above  statement  is 
correct. 
We  shall  now  show  how  to  draw  a  tangent  to  an  ellipse  through 


§71]  THE  ELLIPSE  141 

a  point  S,  not  upon  the  curve.  Let  S,  Fig.  47,  be  a  point  in 
the  plane  of  the  ellipse,  through  which  the  tangent  line  is  to  be 
drawn.  Draw  SS'  parallel  to  PQ,  intersecting  the  plane  OCQ  at 
S'.  Draw  SK  and  S'K  perpendicular  to  BB'  From  the  similar 
right  triangles  SS'K  and  AGO  it  follows  that: 

S'K/SK  =  CO/AO 

Let  ABA'B',  Fig.  54,  be  the  ellipse  to  which  the  tangent  is  to 
be  drawn  from  the  point  S.  Draw  ASW.  Draw  CW.  Draw 
SK  perpendicular  to  BOW,  cutting  CW  at  the  point  S'.  From  S' 
draw  the  two  tangents  to  the  circle  BCB'D.  Let  Qi  and  Q2  be  the 
points  of  tangency.  Draw  RiQiPi  and  RiQ^Pz  perpendicular  to 
B'OB  cutting  the  ellipse  in  the  points  PI  and  P2.  Draw  PiS  and 
PzS.  PiS  and  PyS  are  the  tangents  to  the  ellipse  drawn  from  the 
points  S.  PiS  and  QiS'  intersect  upon  B'BW  ;  and  P*S  and  Q2S' 
intersect  upon  B'BW.  It  is  left  for  the  student  to  show  that  the 
above  statements  are  correct. 

Exercises 

1.  Find  the  value  of  each  semi-axis,  the  focal  distance,  and  the 
eccentricity  for  each  of  the  following: 


(a)  +         =  1-  (c)   lOOz2  +  4i/2  =  400. 
4         y 

(b)  ^  +  !^   =  1.  (d)  7x*  +  3*/2  =  11. 

2.  Construct  an  ellipse  by  the  method  given  in  §  67  if  a   =5 
inches,  and  6=4  inches. 

3.  Construct  an  ellipse  by  the  method  given  in  §  68  if  a  =  4 
and  a  =  30°. 

4.  Find  the  weight  of  metal  removed  in  cutting  the  opening  in  the 
saddle  for  a  circular  ventilator  85  feet  in  diameter,  if  the  rafters  make 
an  angle  of  30°  with  the  horizontal,  and  if  the  ventilator  is  made  of 
20  gauge  iron.     Twenty  gauge  iron  is  0.038  inch  thick,  U.  S.  standard. 
The  specific  gravity  of  steel  is  7.83.     A  cubic  foot  of  water  weighs 
62.5  pounds. 

[5.]  Show  that  a  right  elliptical  cylinder,  has  a  circular  cross  section. 

If, 

x-  +  y-  =  i 

16  ^   9 


142 


MATHEMATICS 


[§71 


M 


is  the  equation  of  its  normal  cross  section,  find  the  angle  between  the 
axis  of  the  cylinder  and  the  plane  giving  the  circular  section. 

[6.]  Show  that  every  section  of  a  right  elliptical  cylinder  is  an 
ellipse. 

[7.]  A  circular  window  in  the  east  wall  of  a  building  is  4.6  feet  in 
diameter.  The  wall  runs  true  north  and  south.  The  rays  of  light 
from  the  sun  pass  through  the  window  and  fall  upon  a  horizontal 
A  floor.  Find  the  area  of  the  floor  in  the  sunlight  if  the 

angle  of  elevation  of  the  sun  is  57°  42'. 6,  and  if  it  is 
37°  27'.7  east  of  the  meridian. 

[8.]  Construct  an  ellipse  with  semi-axes  equal  to 
1^  and  2^  inches.      Draw  a  line  bisecting  the  angle 
between  the  axes.     At  the  point  of  intersection  of 
this    line    with    the    ellipse,    draw  a  tangent  line. 
Measure  the  distances  from  the  center  of  the  ellipse 
to  the  points  of  intersection  of  the  tangent  line  with 
the  axes  of  the  ellipse  produced. 
[9.]    Construct    an    ellipse 
with  semi-axes  equal  to  1\  and 
3|  inches.     Through  its  center 

draw  a  line  making  an  angle  of 

30°  with  the  minor  axis.     From 
a   point  of  this  line  2  inches 
from    the    center   draw    two 
tangents  to  the  ellipse. 

[10.]  Through  the  point  of  tangency  of  the  tangent  line  drawn  in 
exercise  8  draw  a  line  perpendicular  to  the  tangent  line.  With  the 
protractor  measure  the  angles  between  this  perpendicular  and  the 
focal  radii. 

[11.]  Let  OA,  Fig.  55,  be  the  vertical  wall  of  a  building;  OB  a  hori- 
zontal pavement;  MN  a  ladder  of  length  a  +  b.  P  is  a  point  of  the 
ladder  such  that  MP  =  b  and  NP  =  a.  Show  that  as  the  foot  of 
the  ladder  is  drawn  out  along  the  pavement,  P  moves  along  the  arc 
of  an  ellipse. 

HINT:  Consider  OB  as  the  X-axis  and  OA  the  F-axis  of  a  rec- 
tangular coordinate  system.  OT  (  =  x)  and  TP  (  =  y)  are  the  co- 
ordinates of  P.  -  =  sin  a,  and  r  =  cos  a.  Square  these  equations 

d  0 

and  add. 


N 

FIG.  55. 


CHAPTER  V 
[  THE  SLIDE  RULE  ] 

72.  Accuracy  of  Logarithmic  Computation.  The  logarithms 
given  in  tables  are  only  approximations.  In  a  four-place  table, 
they  are  correct  to  the  nearest  fourth  decimal  place;  in  a  five- 
place  table  to  the  nearest  fifth  decimal  place;  in  a  six-place 
table  to  the  nearest  sixth  decimal  place;  etc. 

The  log  ir  =  0.497  to  three  places, 

0.4971          to  four  places, 
0.49715       to  five  places, 
0.497150      to  six  places, 
0.4971499    to  seven  places,  and 
0.49714987  to  eight  places. 

The  log  TT  is  given  correct  to  three,  four,  five,  six,  seven  and 
eight  places  of  decimals;  but  none  of  these  values  is  absolutely 
correct. 

Since  the  logarithms  of  numbers  taken  from  tables  and  used 
in  logarithmic  calculation  are  not  absolutely  correct,  the  results 
of  the  calculations  are  not  absolutely  correct.  But  in  nearly  all 
numerical  calculation,  results  are  required  accurate  only  to  a 
certain  number  of  places;  and  the  number  of  accurate  places  re- 
quired is  the  fact  that  determines  the  table  to  be  used.  If  a  set 
of  logarithms  are  added  (corresponding  to  multiplication)  or 
added  algebraically  (corresponding  to  multiplication  and  division) 
the  errors  in  the  logarithms  will  be,  in  most  cases,  compensating 
in  nature.  That  is  to  say,  the  positive  errors  and  negative  errors 
tend  to  balance,  producing  only  a  small  error  in  the  sum. 

Below  are  written  ten  numbers  selected  at  random.  In  the  sec- 
ond column  are  given  their  logarithms  to  six  places;  in  the  third 
column  their  logarithms  to  four  places;  and  in  the  fourth  column 
the  errors  in  the  four-place  logarithms  as  compared  with  the 
six-place  logarithms. 

143 


144 


MATHEMATICS 


[§72 


Number 

Logarithm 

Logarithm 

Error 

3.72 

0.570543 

0.5705 

-0.000043 

9.81 

0.991669 

0.9917 

+0.000031 

7.15 

0.854306 

0.8543 

-0.000006 

1.23 

0.089905 

0  .  0899 

-0.000005 

5.06 

0.704151 

0.7042 

+0.000049 

7.17 

0.855519 

0.8555 

-0.000019 

8.23 

0.915400 

0.9154 

0.000000 

3.56 

0.551450 

0.5514 

-0.000050 

1.17 

0.068186 

0  .  0682 

+0.000014 

8.75 

0.942008 

0.9420 

-0.000008 

Sum  

6  4431 

-0.000037 

The  sum  of  the  errors  is  —  0.00004,  which  represents  approximately 
the  error  in  the  sum  of  the  ten  logarithms.  The  mantissa  4431 
is  found  to  correspond  to  the  number  2774.  Since  the  tabular  dif- 
ference at  this  point  of  the  table  is  15,  the  above  error,  0.00004 
makes  only  a  difference  of  between  2  and  3  in  the  next  (fifth) 
place  in  the  number.  Thus  by  using  a  four-place  table  in  multi- 
plying these  ten  numbers  together,  the  product  is  found  correct  to 
the  nearest  fourth  place. 

If  the  sum  of  the  ten  logarithms  had  been  6.9872,  the  tabular 
difference  at  this  point  of  the  table  would  have  been  5,  the  error 
of  —  0.00004  would  have  produced  an  error  in  the  corresponding 
number  of  about  7  in  the  fifth  place.  Thus,  the  result  of  the  prod- 
uct of  the  ten  numbers  would  have  been  one  unit  too  small  in  the 
fourth  place. 

We  see  from  this  illustration  that  an  error  in  the  logarithm, 
the  sum,  may  effect  the  fourth  digit  one,  or  possibly  two  points, 
if  the  logarithm  stands  near  the  end  of  the  table;  while  the  fourth 
digit  is  given  correctly  if  the  logarithm  stands  nearer  the  begin- 
ning of  the  table. 

Multiplying  and  dividing  with  a  four-place  table  gives  re- 
sults correct  to  four  places,  or  correct  to  three  places  while  the 
fourth  place  is  in  error  one  or  two  points  only.  The  error  in  the 
product  or  quotient  will  usually  be  less  than  2/100  of  1  percent, 
and  in  the  majority  of  cases  it  will  be  very  much  less  than  this. 


§72]  THE  SLIDE  RULE  145 

This  is  what  is  meant  when  we  say  a  four-place  table  computes  ac- 
curately to  four  places. 

In  taking  out  the  number  corresponding  to  a  logarithm,  do  not, 
in  general,  write  more  than  four  significant  digits  if  a  four-place 
table  is  used.  If  there  are  more  than  four  digits  preceding  the 
decimal  point,  determine  the  first  four,  and  suffix  ciphers  between 
the  fourth  digit  and  the  decimal  point.  Thus,  if  log  x  =  6.1738, 
x  =  1,493,000.  The  fourth  digit  is  determined  by  interpolation, 
and  three  ciphers  are  suffixed.  This  number,  however,  occurring 
near  the  beginning  of  the  table,  could  have  been  corrected  to 
another  place,  in  which  case  the  fifth  digit  would  have  been  correct 
to  within  three  or  four  points. 

In  numerical  work,  use  the  smallest  table  which  will  insure 
the  required  degree  of  accuracy  in  the  result.  If  a  five-place  table 
will  suffice,  do  not  use  a  six-place  table.  If  a  four-place  table 
will  suffice,  do  not  use  a  five-place  table.  If  the  numerical  work 
of  a  problem  is  to  be  done  with  four-place  logarithms,  the  student 
who  has  had  but  little  practice  in  logarithmic  computation  is 
very  apt  to  believe  that  by  using  a  five-place  table  and  discarding 
the  last  digit  he  will  save  the  time  and  trouble  required  in  apply- 
ing corrections.  This  method,  however,  does  not  gain  the  end 
sought;  for,  by  using  a  well-arranged  four-place  table,  corrections 
can  be  applied,  after  a  very  little  practice,  more  expeditiously 
and  easily  than  the  logarithms  can  be  found  from  the  larger  ta- 
ble containing  ten  times  as  many  numbers. 

Another  oversight  on  the  part  of  a  large  number  of  students, 
is  the  attempt  to  compute  more  accurately  than  the  observed 
data  warrant. 

Thus,  to  illustrate,  suppose  a  and  b  are  two  readings.  They  are 
not  absolutely  correct.  The  first  has  a  possible  error  ei  and  the 
second  a  possible  error  e2.  The  two  readings  are  to  be  multiplied 
together.  What  is  the  possible  error  of  the  product?  How  many 
places  in  the  product  are  reliable? 

(a  ±  ci)  (b  ±  e2)  =  ab  ±  ae2  +  6ei  +  ci«2. 

The  sum  of  the  last  three  terms  in  this  product,  taken  with  the 
upper  signs,  is  the  largest  possible  error  in  the  product  due  to 
the  errors  ei  and  e2  in  the  factors  a  and  b,  respectively.     eie2  is 
10 


146  MATHEMATICS  [§73 

small  compared  with  ae2  +  &ei,  and  may  be  neglected.  The  pos- 
sible error  in  the  product  is  then  +  (ac2  +  &ei). 

Suppose  325  and  52 1  are  the  readings.  Suppose  that  in  each  case 
1  in  the  third  place  is  the  least  count  of  the  instrument,  i.e., 
there  is  a  possible  error  of  0.5  in  each  reading.  Then, 

a  =  325 
b  =  521 

€i   =  6z  =   0.5 

ab  =  169,325 
aei  +  6e2  =  423 

Thus  we  see  that  the  product  of  the  two  numbers  whose  measure- 
ments were  attempted  lies  between  169,325  +  423  and  169,325  — 
423.  This  shows  that  the  product  may  be  correct  to  three  places 
only,  and  should  be  written  169,000.  The  digits  following  the  9 
in  the  product  ab  have  no  significance  since  all  may  be  incorrect; 
hence  the  digits  should  not  be  written,  but  ciphers  substituted  in 
their  places. 

In  the  above  illustration,  even  a  four-place  table  carries 
the  accuracy  of  the  work  beyond  what  is  required.  In  this  prob- 
lem the  calculation  could  have  been  done  with  a  three-place 
logarithmic  table,  or  with  a  slide  rule. 

73.  The  Principle  of  the  Slide  Ride.  Let  C  and  D,  Fig.  56, 
represent  two  uniform  scales.  The  two  scales  are  so  placed  that 
the  zero  mark  of  C  stands  opposite  the  3  mark  of  D.  Any  number 
on  D  to  the  right  of  3  is  the  sum  of  the  number  directly  above  it  on 
C,  and  3.  Thus,  5(read  on  scale  D)  is  the  sum  of  2  and  3;  6  is  the 
sum  of  3  and  3;  4|  is  the  sum  of  1J  and  3;  etc.  In  this  way 
any  number  may  be  added  mechanically  to  the  number  3.  By 
placing  the  0  of  C  above  any  point  of  D,  any  other  number  may 
be  added  mechanically  to  the  number  corresponding  to  this  point 
of  the  D  scale. 

It  is  immaterial  whether  the  0  of  C  or  the  0  of  D  is  set  to  cor- 
respond to  one  of  the  numbers  to  be  added.  If  the  0  of  D  is  set, 
the  sum  is  read  from  the  C  scale. 

By  a  similar  process,  subtraction  may  be  performed  mechanic- 
ally. Thus,  if  3  is  to  be  subtracted  from  6,  move  scale  C  so  that 


§73]  THE  SLIDE  RULE  147 

3  upon  it  stands  above  6  of  D,  the  number  of  D  below  the  0  of  C, 
or  3,  is  the  difference. 

In  exercise  10,  §  41,  there  were  constructed  two  logarithmic 
scales,  each  in  juxtaposition  with  a  uniform  scale.1  These  two 
pairs  of  scales  are  represented  (reduced)  in  Fig.  57.  Bend  scales 
C  and  D  along  the  horizontal  lines;  turn  the  lower  part,  the  uniform 
scale,  of  C  back  behind  the  non-uniform  scale;  and  turn  the 
upper  part  of  D  back  behind  the  lower  part.  Now  place  these  two 
scales  together  as  shown  in  Fig.  58.  The  lines  marked  1  and  10  of 
these  scales  are  called  indices. 

If  we  bring  the  left-hand  index  of  C  opposite  2  of  D,  any  number 
to  the  right  of  2  on  the  D  scale  will  be  the  product  of  the  corre- 
sponding number  of  C,  and  2.  Thus,  in  the  figure,  6  of  D  stands 
beneath  3  of  C;  6  is  the  product  of  3  and  2.  To  multiply  2  by 
3  mechanically,  bring  the  index  of  the  upper  scale  over  2  of  the 
lower  scale,  and  read  the  number  on  the  lower  scale  which  stands 
directly  under  3  of  the  upper  scale.  This  number  is  the  product. 
For,  upon  the  back  of  each  scale,  C  and  D  is  a  uniform  scale, 
and  by  sliding  C  and  D  upon  each  other  we  are  adding  me- 
chanically, as  was  shown  above,  the  numbers  of  these  uniform 
scales.  Thus,  on  the  back  of  D  is  a  number,  which  is  the  sum 
of  the  number  back  of  2  on  D  and  the  number  back  of  3  on  C. 
But,  the  number  back  of  2  on  D  is  the  logarithm  of  2,  and  the 
number  back  of  3  on  C  is  the  logarithm  of  3;  then  on  the  back 
of  D  opposite  3  on  C  is  the  logarithm  of  2  plus  the  logarithm  of 
3,  or  the  logarithm  of  6,  since  the  sum  of  the  logarithms  of  two 
numbers  is  the  logarithm  of  their  product.  On  the  face  of  D  are 
numbers  whose  logarithms  are  on  the  back.  Therefore,  opposite 
the  logarithm  of  6  upon  the  back  of  D,  i.e.,  opposite  3  of  C,  we 
have  6,  or  the  product  of  2  and  3. 

In  constructing  scales  C  and  D,  one  unit  of  length  only,  of  the 
uniform  scale  was  used.  By  extending  these  double  scales  indefi- 
nitely in  both  directions,  the  uniform  parts  will  include  all  al- 
gebraic numbers,  i.e.,  all  numbers  from  —  °°  to  +  <»  ;  the  non-uni- 
form parts  will  include  all  positive  arithmetical  numbers,  i.e.,  all 
numbers  from  0  to  +  °° .  A  port:on  of  one  of  these  scales  (the  C 

i  If  this  exercise  was  not  performed  when  the  work  on  logarithms  was  taken  up,  it 
should  be  performed  now,  or  at  least  the  description  read  carefully. 


148 


MATHEMATICS 


[§73 


O          -  :  :  - 


§73]  THE  SLIDE  RULE  149 

scale)  is  represented  in  Fig.  59.  The  extended  D  scale  is  the  same, 
but  inverted.  For  convenience  of  description,  different  segments 
of  these  scales  will  be  referred  to  as  "block  I,"  "block  II,"  etc. 
By  block  II  is  meant  that  portion  of  the  double  scale  between  1  and 
2  of  the  uniform  part,  or  between  10  and  100  of  the  non-uniform 
part;  by  block  —  I  is  meant  that  portion  of  the  uniform  part 
between  —  1  and  0,  or  between  0.1  and  1  of  the  non-uniform 
part,  etc. 

Following  directly  from  the  facts  that  the  mantissas  of  the  com- 
mon logarithms  are  independent  of  the  position  of  the  decimal 
point  in  the  number,  and  that  by  moving  the  decimal  point  one 
place  to  the  right  or  left  increases  or  decreases,  respectively, 
the  characteristic  of  the  logarithm  by  1,  it  is  seen  that  all  blocks 
of  the  double  scale,  Fig.  59,  are  identical.  To  illustrate:  the  line 
on  the  non-uniform  part  representing  20  is  in  the  same  position 
relative  to  10  as  the  line  representing  2  is  to  1.  This  must  be  so, 
for  the  logarithm  of  20  equals  the  logarithm  of  10  plus  the  loga- 
rithm of  2;  and,  since  the  logarithms  are  represented  on  the  uniform 
part,  the  logarithm  of  20  is  as  far  to  the  right  of  1  on  the  uniform 
part  as  the  logarithm  of  2  is  to  the  right  of  0.  Since  all  blocks 
are  the  same,  one  block  will  suffice  for  the  entire  indefinite  scale. 
In  using  scales  C  and  D,  Fig.  58,  for  mechanical  multiplication 
the  operator  may  disregard  the  positions  of  the  decimal  points  in 
the  factors,  and  point  off  the  proper  number, of  places  in  the 
product  after  the  sequence  of  digits  has  been  found. 

Since  the  index  at  the  right-hand  end  of  C  may  be  looked  upon 
as  1  as  well  as  10,  the  right-hand  index  as  well  as  the  left-hand 
index  may  be  set  above  one  of  the  factors  of  the  product.  It 
will  be  seen,  however,  that  in  multiplying  two  numbers  together, 
using  only  one  block  each  of  the  C  and  D  scales,  only  one  of  the 
two  indices  of  C  can  be  set  over  the.mutiplicand,  for  if  the  other 
index  were  set  it  would  throw  the  product  off  the  D  scale  into 
the  next  block.  By  interchanging  the  multiplicand  and  multiplier 
the  other  index  of  C  may  be  set.  To  illustrate:  let  3  be  multi- 
plied by  6.  Set  the  left-hand  index  of  C  over  3,  then  6  of  C 
falls  beyond  the  D  scale.  In  this  case  the  product  may  be  found 
either  by  setting  the  right-hand  index  of  C  over  3  and  reading 


150 


MATHEMATICS 


[§74 


cq  O  C) 


under  6,  or  by  setting  the  right-hand  index 
of  C  over  6  and  reading  under  3. 

By  means  of  scales  C  and  D  any  two  num- 
bers may  be  multiplied  together  mechan- 
ically. The  principle  involved,  i.e.,  of 
adding  logarithms  mechanically,  is  the  un- 
derlying principle  of  the  construction  of 
the  (Mannheim)  slide  rule.1  The  opera- 
tions performed  by  the  mechanism  may  be 
compared  with  logarithmic  work  as  follows. 
Locating  the  numbers  on  the  scales  may  be 
compared  with  taking  out  the  logarithms  of 
the  numbers  and  the  anti-logarithm  of  their 
product,  while  sliding  the  C  scale  above  the 
D  scale  may  be  compared  to  the  addition 
^  of  the  logarithms. 

2  Since  the  logarithms  of  the  factors  and 

3  the  logarithm  of  their  product  are  not  read 
""  they  need  not  be  printed  upon  the  back  of 
^  the  scales  C  and  D. 

|       74.  The  Slide  Rule.     The  slide  rule  con- 

§  sists  of  three  parts:  the  body,  the  slide,  and 

e  the  indicator  (Fig.  60).     The  body  of  the  rule 

£  consists  of  a  wooden  base  about  10  inches 

long,2  about  1  inch  wide,    and  about   1/8 

inch    thick,    upon    one    side  of  which  are 

fastened  two  parallel  strips,  each  carrying 

1  Many  mechanical  devices  have  been  constructed  for 
performing  special  numerical  operations.  Often  they  are 
in  the  form  of  one  scale  sliding  over  another,  and  for 
•  this  reason  are  also  termed,  with  a  proper  qualifying 
adjective, -slide  rules;  as,  stadia  slide  rule,  sewer  slide 
rule,  horse-power  slide  rule,  etc. 

The  slide  rule  is  also  constructed  by  placing  the 
logarithmic  scales  upon  two  concentric  circular  arcs,  one 
rotating  within  the  other. 

'  Slide  rules  are  made  in  three  sizes:  viz.,  the  5-inch, 
about  5  inches  long;  the  10-inch,  about  10  inches  long; 
and  the  20-inch,  about  20  inches  long.  The  10-inch  is 
the  size  commonly  used.  The  5-inch  is  less  accurate, 
but  is  very  convenient  for  carrying  in  the  pocket  and 
for  rough  estimates. 


§75]  THE  SLIDE  RULE  151 

a  logarithmic  scale  engraved  upon  celluloid.  These  two  scales  are, 
in  Fig.  60a,  marked  A  and  D.  A  transverse  section  of  the  rule  is 
shown  in  Fig.  606.  The  two  upper  parallel  strips  of  the  body 
are  separated  about  1/2  inch,  and  between  them  there  is  placed 
the  slide,  a  thin  strip  of  wood  carrying  two  logarithmic  scales, 
B  and  C.  The  scales  C  and  D  are  the  same  as  those  explained 
in  the  preceding  section,  and  by  means  of  them,  multiplications 
may  be  performed. 

The  indicator  /,  Fig.  60a,  consists  of  a  piece  of  glass,  about  an 
inch  square,  mounted  in  an  aluminum  frame,  which  slides  in 
grooves  lengthwise  the  rule.  On  the  under  side  of  the  glass, 
very  nearly  in  contact  with  the  surfaces  of  the  four  scales,  is  etched 
a  fine  vertical  line,  or,  "cross  line"  as  it  will  be  called. 

Scales  A  and  B  are  logarithmic  scales  exactly  one-half  as  long 
as  scales  C  and  D,  i.e.,  upon  A  and  B  are  two  blocks  of  the  in- 
definite logarithmic  scale.  Scales  A  and  B,  then,  have  each 
three  indices,  one  at  each  end  and  one  at  the  mid-point.  When  we 
come  to  the  explanation  of  the  use  of  the  A  and  B  scales  we  shall 
see  that  the  end  indices  are  considered  to  correspond  to  the  num- 
bers .  .  .  0.0001,0.01,1,100,10,000,.  .  .  ,  while  the  central  in- 
dex corresponds  to  the  numbers  .  .  .  0.001,  0.1,  10,  1000, 
100,000.  .  . 

The  scales  A  and  B  are  placed  upon  the  body  of  the  rule 
so  that  the  left-hand  index  of  A  is  directly  over  the  left-hand 
index  of  D,  and  so  that  the  right-hand  index  of  A  is  directly 
over  the  right-hand  index  of  D.  When  the  cross  line  of  the  indi- 
cator is  set  over  the  left-hand  index  of  A  it  will  also  be  over  the 
left-hand  index  of  D;  when  set  over  the  right-hand  index  of  A  it 
will  also  be  over  the  right-hand  index  of  D. 

75.  The  multiplication  of  two  or  more  numbers  will  be  il- 
lustrated by  an  example.  To  multiply  (37.5)  (212)  (11.2)  (0.315) 
proceed  as  follows:  Set  left-hand  index  of  C  over 375  of  D.  Set 
cross  line  of  indicator  over  212  of  C.  (Directly  under  the  cross  line 
on  the  D  scale  is  the  product  of  the  first  two  factors,  but  do  not 
read  it.)  Set  the  left-hand  index  of  C  under  cross  line  (i.e.,  over 
the  product  of  the  first  two  factors  on  Z>).  Set  cross  line  over 
112  on  C.  Bring  the  right-hand  index  of  C  under  the  cross  line. 
Move  cross  line  over  315  of  C,  and  under  it  on  the  D  scale  read 


152  MATHEMATICS  [§76 

the  product  of  the  four  factors.  The  digits  read  are  2805.  To 
locate  the  decimal  point,  find  mentally  a  very  rough  estimate 
of  the  product;  thus,  the  product  of  the  first  two  factors  is  about 
i  of  21,000,  or  about  7000;  the  next  multiplier  (about  10) 
makes  the  product  of  the  first  three  factors  about  70,000;  the 
next  multiplier  is  about  ^,  which  makes  the  product  of  four 
factors  about  23,000.  This  shows  that  there  are  five  digits  be- 
fore the  decimal  point.  The  product  is  then  28,050. 

It  is  to  be  noticed  that  in  the  above  operations  no  readings 
are  taken  from  the  rule  excepting  the  final  product. 

In  a  similar  way  any  number  of  factors  may  be  multiplied 
together. 

If  several  numbers  are  to  be  multiplied  by  the  same  number, 
set  the  index  of  C  over  the  common  factor  on  D.  Slide  in  turn  the 
cross  line  over  the  given  numbers  on  C,  and  read  off  the  various 
products  from  D.  In  this  way  all  the  products  can  be  read  with 
one,  or  at  most  two,  settings  of  the  slide. 

76.  Division.  The  division  of  one  number  by  another  may  be 
performed  upon  a  slide  rule  in  two  ways.  The  principle  of  one 
method  is  that  subtracting  the  logarithm  of  the  divisor  from  the 
logarithm  of  the  dividend  gives  the  logarithm  of  the  quotient. 
The  principle  of  the  other  method  is  that  multiplying  the  quotient 
by  the  divisor  gives  the  dividend.  Both  methods  will  be  illus- 
trated by  example.  To  divide  372  by  176,  set  the  slide  so  that 
176  of  C  is  directly  above  372  of  D.  (To  do  this  easily,  first  set  the 
cross  line  over  372  of  D  and  then  bring  176  of  C  under  it.)  Read 
the  quotient  from  D,  the  number  under  the  index  of  C.  The 
quotient  is  2.113.  By  inspection  it  is  seen  that  the  decimal  point 
follows  the  2. 

This  method  of  division  corresponds  to  the  subtraction  of 
logarithms.  For,  the  distance  from  the  index  to  372  (measured 
upon  the  uniform  scale  on  the  back  of  D)  is  the  logarithm  of  372. 
The  distance  from  the  index  of  C  to  the  number  176  is  the  logarithm 
of  176.  Then  the  distance  from  the  index  of  D  to  the  index  of  C 
is  the  difference  of  these  two  logarithms,  or  the  logarithm  of  the 
quotient;  and  the  number  on  the  face  of  D  is  then  the  quotient. 

To  divide  by  the  second  method,  set  the  index  of  C  over  the 
divisor,  176,  on  D.  Set  the  cross  line  over  the  dividend,  372,  on 


§77]  THE  SLIDE  RULE  153 

D.  Under  the  cross  line  on  the  C  scale  read  the  quotient,  2. 1 13 
It  is  easily  seen  that  this  method  gives  the  quotient,  for  it  is  the 
reverse  of  that  used  in  multiplication.1 

The  second  method  of  division  is  useful  if  several  numbers  are 
to  be  divided  by  a  common  number;  for  example,  to  find  the  per- 
centages which  a  set  of  numbers  are  of  a  given  number,  place  the 
index  of  C  over  the  common  divisor  on  D,  bring  the  cross  line  over 
the  dividend  on  D  and  read  the  quotient  from  C  under  the  cross 
line.  In  this  way  a  series  of  quotients  may  be  found  with  one,  or 
at  most  two,  settings  of  the  slide. 

77.  The  A  and  B  Scales.     Since  the  "block"  on  A  is  just  one- 
half  as  long  as  the  "block"  on  D,  the  numbers  (in  imagination)  on 
the  back  of  A  are  just  double  those  directly  opposite  on  the  back 
of  D.    The  numbers  on  the  face  of  A  are  then  the  squares  of  the 
numbers  directly  opposite  on  the  face  of  D.    This  follows  from 
the  fact  that  the  logarithm  of  the  square  of  a  number  is  two  times 
the  logarithm  of  the  number. 

ILUSTRATIONS :  To  find  the  square  of  3.77,  either  multiply 
3.77  by  3.77  or  set  the  cross  line  over  3.77  on  the  D  scale  and 
read  its  square,  14.2,  from  the  A  scale.  To  extract  the  square 
root  of  7.13,  set  the  cross  line  over  713  of  the  left-hand  block 
of  the  A  scale,  read  off  its  square  root,  2.67,  from  the  D  scale. 
To  extract  the  square  root  of  71.3,  set  the  cross  line  over  713  of 
the  right-hand  block  of  A,  read  off  its  square  root,  8.44,  from  the 
D  scale. 

In  extracting  square  roots  care  must  be  exercised  to  enter  the 
number  in  the  proper  block  of  the  A  scale. 

Scales  A  and  B  may  be  used  for  multiplication  and  division 
in  the  same  way  that  scales  C  and  D  are  used.  But  on  account  of 
the  A  and  B  scales  being  only  half  as  long,  the  results  are  not  as 
accurate  as  when  C  and  D  are  used.2 

78.  The  Cube  Root.3    The  method  of  finding  cube  roots  of 
numbers  will  be  illustrated  by  examples. 

1  The  first  method  may  also  be  looked  upon  as  the  reverse  of  multiplication. 

1  The  Polyphase  Slide  Rule  is  one  having  two  additional  scales.  One  of  these 
scales  gives  cubes  (and  cube  roots)  directly;  the  other  gives  reciprocals  of  numbers 
directly. 

*  Some  rules  are  provided  with  a  cube  scale  by  means  of  which  cubes  and  cube  roots 
may  be  obtained  directly. 


154  MATHEMATICS  [§79 

Example  1 :  Find  the  cube  root  of  8.  Set  the  cross  line  over 
8  on  the  A  scale  (block  I).  Move  the  slide  so  that  the  number 
on  the  B  scale  under  the  cross  line  is  the  same  as  the  number  on 
the  D  scale  under  the  left-hand  index  of  the  C  scale.  This  is 
found  to  be  2,  the  cube  root  of  8. 

Example  2:  Find  the  cube  root  of  80.  Set  the  cross  line  over 
8  on  the  A  scale  (block  I).  Move  the  slide  so  that  the  number  on 
the  B  scale  under  the  cross  line  is  the  same  as  the  number  on  the 
D  scale  under  the  right-hand  index  of  the  C  scale.  This  is  found 
to  be  4.31,  the  cube  root  of  80. 

Example  3:  Find  the  cube  root  of  800.  Set  the  cross  line 
over  8  on  the  A  scale  (block  II).  Move  the  slide  so  that  the 
number  on  the  B  scale  under  the  cross  line  is  the  same  as  the 
number  on  the  D  scale  under  the  right-hand  index  of  the  C  scale. 
This  is  found  to  be  9.28,  the  cube  root  of  800. 

The  student  will  explain  why  these  operations  give  the  cube 
roots. 

79.  Other  Operations  with  the  Slide  Rule.  It  was  the  inten- 
tion to  give  above  only  the  underlying  principle  of  the  slide 
rule;  and  to  explain  its  use  in  connection  with  the  simple  opera- 
tions of  multipli  cation,  division,  extracting  square  root,  and 
squaring.  It  is  believed  that  the  student  will,  after  some  prac- 
tice with  the  slide  rule,  devise  methods  of  procedure  which  will 
involve  the  fewest  number  of  settings  of  the  slide,  and  the  fewest 
number  of  intermediate  readings.  Pamphlets  and  books  may 
be  procured  giving  in  considerable  detail  the  methods  of  perform- 
ing numerical  calculation  with  the  slide  rule. 

In  performing  numerical  calculations  with  the  slide  rule,  the 
work  should  be  blocked  out  as  when  calculating  with  logarithms.1 

Exercises 

The  student  will  practice  on  simple  multiplications,  divisions, 
squares,  square  roots,  and  cube  roots. 

1  Paper,  form  M 8,  is  ruled  especially  for  slide  rule  calculations. 


CHAPTER  VI 
STATICS 

80.  Vector  and  Scalar   Quantities.     In  Physics  and  in  Me- 
chanics we  have  to  deal  with  two  kinds  of  quantities:  vector 
quantities  and  scalar  quantities. 

Quantities,  such  as  velocity,  force,  acceleration,  etc.,  which 
require  for  their  complete  definition  direction  as  well  as  magnitude, 
are  called  vector  quantities.  When  we  say  a  body  has  a  uniform 
velocity  of  10  feet  per  second,  we  mean  that  in  every  second  of 
time  the  body  moves  a  distance  of  10  feet  and  in  some  given 
direction.  Velocity,  then,  is  a  vector  quantity. 

Quantities,  such  as  mass,  time,  area,  volume,  etc.,  having 
only  magnitude,  are  called  scalar  quantities.  In  order  to  convert 
a  gram  of  ice  at  zero  degrees  centigrade  into  water  at  the  same 
temperature,  79  units  of  heat  must  be  added  to  the  mass.  This 
quantity,  79  calories,  is  a  scalar  quantity. 

81.  Graphic  Representation  of  Vector  Quantities.     In  order 
to  represent  a  vector  quantity  graphically,  choose  some  unit  of 
length,  as  the  inch,  the  half-inch,  the  centimeter,  to  represent 
the  unit  of  magnitude;  with  this  unit  length,  draw  a  line  whose 
length  measures  the  magnitude  of  the  quantity,  and  whose  direc- 
tion is  that  of  the  vector  quantity.     To  illustrate:  a  body  moves 


FIG.  61. 

from  west  to  east  with  a  speed  of  3  miles  per  hour.  Draw  the  line 
AB,  Fig.  61,3  cm.  long  and  from  left  to  right.  The  line  drawn 
3  cm.  long  represents  the  magnitude,  the  speed,  3  miles  per  hour, 
and  drawn  from  A  to  B  represents  the  motion  from  left  to  right, 
or,  from  west  to  east.  An  arrow  is  placed  upon  the  line  indicat- 
ing its  sense,  i.e.,  from  left  to  right. 

A  directed  line,  as  AB,  Fig.  61,  is  called  a  vector. 

155 


156  MATHEMATICS  [§82 

A  vector  quantity  is  also  termed,  for  brevity,  simply  a  vector. 

In  what  follows  it  is  assumed  that  all  problems  are  of  such  a 
nature  that  all  vector  quantities  act  in  the  same  plane.  The  vec- 
tors representing  these  vector  quantities  may  then  be  drawn  upon  a 
plane,  the  plane  of  the  paper. 

82.  Vector  Addition.  In  elementary  algebra  real  numbers  are 
usually  represented  graphically  by  points  along  a  straight  line. 
Thus,  if  0,  Fig.  62,  be  any  point  upon  the  straight  line  X'X, 


FIG.  62. 

points  to  the  right  of  0  represent  the  positive  numbers,  while 
points  to  the  left  represent  the  negative  numbers.  To  every  real 
number  there  corresponds  some  point  upon  the  line,  and  for  every 
point  of  the  line  there  exists  a  real  number.  If  the  points  upon 
the  line  are  chosen  such  that  distances  between  them  are  propor- 
tional to  the  steps  between  the  numbers  which  they  represent, 
we  have  what  was  previously  called  a  uniform  scale.  In  what 
follows  this  uniform  scale  is  assumed  as  the  graphic  representation 
of  the  real  number  system.  The  points  A,B,C,  D,  etc.,  or  the 
vectors  OA,  OB,  OC,  OD,  etc.,  represent,  respectively,  the 
numbers  2,  5,  -3,  -4,  etc. 

o  A      B  c 

X  -  1  -  1  ____  i  ___  >!  ___  >!  ____  i  ___  J  -  X 


FIG.  63. 

The  addition  of  two  real  numbers,  say  2  and  3,  may  be  looked 
upon  as  consisting  of  the  following  operations:  Move  the  vector 
OA,  Fig.  63,  parallel  to  itself;  place  it  so  that  its  initial  point, 
which  was  originally  at  0,  coincides  with  the  terminal  point, 
B,  of  the  vector  OB.  Its  new  position  is  indicated  by  the  dotted 
line  BC.  Draw  the  vector  OC.  It  is  easily  seen,  since  the  scale 
X'X  is  uniform,  that  the  vector  OC  represents  the  number  5,  the 
sum  of  3  and  2.  It  is  readily  seen  that  this  process  of  addition 
holds  for  any  set  of  real  numbers  (negative  as  well  as  positive), 
and  that,  in  the  process,  either  of  the  two  vectors  may  be  moved. 


§82] 


STATICS 


157 


In  general,  let  OB  and  OA,  Fig.  64,  be  any  two  vectors.  Move 
OA  parallel  to  itself  so  that  its  initial  point  coincides  with  the 
terminal  point,  B,  of  the  vector  OB.  It  then  takes  up  the  po- 
sition BC.  Draw  the  vector  OC.  The  vector  OC  is  defined  as  the 
vector  sum  of  the  vectors  OB  and  OA. 

The  negative  of  a  vector  is  a  vector  of  the  same  length,  but 
extending  in  the  opposite  direction  (having  opposite  sense). 


FIG.  64. 

« 

To  subtract  the  vector  OA  from  the  vector  OB  means  to  find 
a  third  vector  OC,  Fig.  65,  such  that  the  sum  of  the  vectors 
OC  and  OA  is  the  vector  OB.  If  OA  is  moved  parallel  to  itself,  so 
that  its  terminal  point,  A,  falls  upon  the  point  B,  and  its  initial 
point  at  C,  the  vector  OC  is  such  that  the  vector  OC  plus  the  vector 
OA  equals  the  vector  OB.  Therefore, 

vector  OC  =  vector  OB  —  vector  OA 


FIG.  65. 

From  the  figure  it  is  readily  seen  that,  to  subtract  the  vector  OA 
from  the  vector  OB  is  to  add  the  negative  of  the  vector  OA ,  that 
is,  AO,  to  the  vector  OB. 

To  find  graphically  the  vector  sum  of  any  three  vectors,  find 
the  sum  of  any  two  and  to  this  sum  add  the  third. 


158 


MATHEMATICS 


[§83 


A  vector  is  designated  by  giving  its  length,  and  the  angle 
which  it  makes  with  the  positive  direction  of  the  axis  of  x.  Thus 
(10,  30°)  when  referring  to  a  vector  means  that  the  vector  is  ten 
units  long  and  its  direction  makes  an  angle  of  30°  with  the  positive 
direction  of  the  axis  of  x. 


Exercises 

Find  graphically  the  sum  of  the  following  vectors: 

1.  (2,  10°)  and  (3,  20°).  3.  (6,  70°)  and  (5,  217°). 

2.  (5,  30°)  and  (10,  135°).  4.  (5,  110°)  and  (3,  -  71°). 
6.  (3,  30°),  (7,  121°)  and  (11,  -  75°). 

83.  Resultant   Velocity.    Let  a  raft   move   along  a  straight 
path  with  uniform  speed  of  2  feet  per  second;  suppose  a  man 


Direction  of  Raft 


FlG.   66. 

upon  the  raft  walks  at  the  rate  of  3  feet  per  second  in  the  direc- 
tion indicated  in  Fig.  66.  Let  A  be  the  point  of  the  raft 
occupied  by  the  man  at  some  particular  instant.  In  one  second 
of  time  the  point  A  "will  be  translated  by  the  movement  of  the  raft 
to  the  point  B.  The  length  of  the  line  AB  represents  2  feet, 
and  the  direction  of  AB  is  the  direction  of  motion  of  the  raft.  If 


§84]  STATICS  159 

the  raft  were  not  moving,  the  man  would  walk  along  the  line  A  C, 
and  at  the  end  of  one  second  would  be  at  the  point  C.  The  motion 
of  the  raft,  however,  has  carried  the  point  C  to  the  point  D ;  at  the 
end  of  one  second  of  time  the  man  is  at  the  point  D  instead  of  at 
the  point  C.  ABDC  is  a  parallelogram.  The  sides  AB  and  AC 
are  vectors  representing,  respectively,  the  velocity  of  the  raft 
relative  to  the  earth  and  the  velocity  of  the  man  relative  to  the 
raft.  By  the  same  reasoning  it  is  seen  that  at  the  end  of  the 
second  second  the  man  will  be  at  the  point  G.  The  points  A,  D, 
and  G  are  upon  a  straight  line.  At  the  end  of  the  first  half 
second  the  man  was  at  the  point  K,  the  center  of  the  parallelogram 
ABDC,  and  at  the  end  of  any  other  fractional  part  of  one  second  he 
was  at  some  point  X  on  the  line  AD.  From  similar  triangles,  it 
is  easily  shown  that  X  is  not  only  on  the  line  AD,  but  at  a  position 
such  that  AX  is  to  A  D  as  the  time  interval  is  to  one  secohd. 
Thus,  the  man  moves,  relative  to  the  earth,  in  the  direction 
AD,  with  a  uniform  speed,  and  over  a  distance  AD  in  one  second 
of  time.  This  is  only  another  way  of  saying  that  the  velocity 
of  the  man  may  be  represented  by  the  vector  sum  of  the 
vector  representing  the  velocity  of  the  raft  relative  to  the  earth, 
and  the  vector  representing  the  velocity  of  the  man  relative  to  the 
raft.  In  general,  then,  if  a  body  is  subjected  to  several  velocities, 
its  resultant  velocity  may  be  found  by  taking  the  vector  sum  of 
the  vectors  representing  the  component  velocities. 

84.  Resultant  of  Concurrent  Coplanar  Forces  Found  Graphic- 
ally. Experimentally,  it  is  shown  that  the  resultant  of  two 
concurrent1  forces  is  such  that  it  may  be  found  in  magnitude  and 
direction  by  considering  it  the  vector  sum  of  the  two  vectors 
representing  the  two  given  forces.  If  a  force  of  10  pounds  acts 
horizontally  to  the  right,  and  if  a  second  force  of  6  pounds  acts  in 
a  direction  making  an  angle  of  60°  with  the  horizontal,  the  result- 
ant of  the  two  may  be  found  graphically  in  the  following  manner. 
Lay  off  a  line  AB  5  inches  long,  Fig.  67.  The  vector  AB  repre- 

1  Forces  acting  upon  a  body  are  concurrent  if  they  have  the  same  point  of  applica- 
tion; otherwise  they  are  non-concurrent. 

Forces  acting  upon  a  body  are  coplanar  if  their  lines  of  action  lie  in  a  plane;  other- 
wise they  are  non-coplanar. 

A  single  force,  if  it  exists,  which  is  equivalent  to  a  system  of  forces  is  called  the 
resultant  of  the  system. 


160 


MATHEMATICS 


sents  the  10-pound  force.  One  inch  in  length  represents  a  magni- 
tude of  two  pounds.  Draw  the  line  AC  3  inches  long  making  the 
angle  BAG  60°.  The  vector  AC  represents  the  6-pound  force. 
Complete  the  parallelogram  ABDC,  or  the  triangle  ABD,  or  the 


Reduced  B 

FIG.  67. 

triangle  ACD.  AD  is  a  vector  representing  the  resultant  of  the  10- 
and  the  6-pound  forces.  Measuring  the  length  of  AD  it  is  found 
to  be  7  inches,  which  shows  that  the  magnitude  of  the  resultant 
force  is  14  pounds.  Measuring  the  angle  BAD  it  is  found  to  be 
approximately  21°  45',  the  direction  angle  of  the  resultant. 
ABDC  is  called  the  parallelogram  of  forces. 


Reduced 


FlG.    68. 


The  resultant  of  three  or  more  concurrent  coplanar  forces  may  be 
found  by  repeated  application  of  the  parallelogram  of  forces. 
Thus,  to  find  the  resultant  of  the  five  forces,  PI,  P%,  Pz,  Pt,  PS, 
whose  magnitudes  are,  respectively,  3,  6,  2,  2,  and  4  pounds,  and 


§85]  STATICS  161 

whose  directions  make,  respectively,  the  angles  0°,  30°,  70°,  210° 
and  170°  with  the  positive  direction  of  the  axis  of  x,  proceed  as 
follows:  Draw  the  vector  AB,  Fig.  68,  3  inches  long  in  a  hori- 
zontal direction.  Draw  BC  6  inches  long  making  an  angle 
of  30°  with  the  horizontal.  AC  is  a  vector  representing  the  result- 
ant of  the  forces  PI  and  P2.  Draw  a  vector  CD  representing  the 
force  P3.  AD  is  a  vector  representing  the  resultant  of  PI,  P2  and 
P3.  Draw  the  vector  DE  representing  the  force  P4.  AE  is  the 
vector  representing  the  resultant  of  PI,  P2,  PS,  and  P4.  Draw 
EF,  a  vector  representing  the  force  P6.  AF  is  the  vector  repre- 
senting the  resultant  of  PI,  P2,  PS,  P4,  and  P5.  AF  is  5.59  inches 
in  length,  which  shows  that  the  magnitude  of  the  resultant  force  is 
5.59  pounds.  The  angle  BAF  measured  with  a  protractor  is 
found  to  be  approximately  55°. 

It  is  seen  that  in  the  construction,  the  lines  AC,  AD,  and  AE 
are  entirely  unnecessary  and  should  be  omitted.  The  figure 
ABCDEFA  is  called  the  polygon  of  forces.  AF,  the  resultant,  is 
called  the  closing  side.  It  will  be  noticed  that  the  arrows  on  the 
vectors  representing  the  given  forces  all  run  in  the  same  sense 
around  the  polygon,  while  the  arrow  of  the  closing  vector  runs  in 
the  opposite  sense. 

If  the  vectors  representing  the  given  concurrent  forces  of  them- 
selves form  a  closed  polygon,  the  resultant  is  zero,  or  the  system  of 
given  forces  is  in  equilibrium. 

Exercises 

Find  graphically  the  resultant  of  the  following  concurrent,  coplanar 
forces : 

1.  (3,  30°),  (5,  45°),  and  (7,  190°). 

2.  (2,  0°),  (3,  40°),  (5,  170°),  and  (7,  135°). 

3.  (2,  -  10°),  (6,  90°),  (5,  0°),  and  (7,  135°). 

4.  (3,  180°),  (2,  270°),  (5,  45°),  and  (1,  150°). 

5.  (1,  270°),  (5,  -  30°),  (7,  90°),  and  (2,  30°). 

85.  Resultant  of  Concurrent  Coplanar  Forces  found  Analytic- 
ally. Let  AD,  Fig.  69,  represent  a  force  both  in  magnitude  and 
direction.  If  AB,  be  a  line  of  any  length,  drawn  in  any  direction, 
and  if  ABiDCi  be  a  parallelogram  so  constructed  that  ABi  is  a  side 
and  AD  a  diagonal,  AD  may  be  considered  the  resultant  of  two 
11 


162 


MATHEMATICS 


[§85 


forces  represented  by  AE\  and  ACi.  Then  in  a  process  of  analysis 
the  force  represented  by  AD  may  be  replaced  by  the  two  forces 
represented  by  ABi  and  AC\.  The  force  AD  is  said  to  be  resolved 
into  the  components  AB\  and  ACi.  It  is  apparent  that  AD  can 
be  resolved  into  an  infinite  number  of  pairs  of  components;  and 

if  it  were  desirable, 
either,  or  both,  of 
these  components 
could  be  resolved 
further  into  two  or 


A  BI          B 

•pIG   gg  more  components. 

If  the  resolution  is 

such  that  the  components  are  parallel  one  to  each  axis  of  coordi- 
nates, the  force  is  said  to  be  resolved  into  its  x-  and  ^-com- 
ponents, or  into  its  horizontal  and  vertical  components. 

It  is  seldom  desirable  to  resolve  a  force  into  other  than  its  x- 
and  ^-components,  and  this  form  of  resolution  is  very  helpful  at 
times  in  finding  the  resultant  force  acting  upon  a  body. 

If  OD,  Fig.  70,  represents  a  force  acting  in  a  direction  making 
an  angle  6  with  the  positive  direction  of  the  axis  of  x,  the  magni- 
tudes of  the  horizontal  and  vertical  components  are,  respectively, 
R  cos  6  and  R  sin  B,  where  R  is  v 
the  magnitude  of  OD.  The 
vector  representing  the  force  in 
Fig.  70  is  in  the  first  quadrant; 
but  the  student  will  easily  see, 
by  sketching  other  figures,  that 
the  algebraic  sign  of  the  sin  6  and 
cos  6  will  give  to  the  component 
force  the  proper  algebraic  sign, 
i.e.,  positive  when  up  or  to  the 
right,  and  negative  when  down  or  to  the  left. 

What  is  here  called  the  analytic  method  of  finding  the  resultant 
of  a  set  of  given  forces,  consists  of  the  following  process:  resolve 
each  force  into  its  horizontal  and  vertical  components,  add  alge- 
braically the  horizontal  components,  add  algebraically  the  vertical 
components,  and  from  these  two  sums  find  the  resultant  as  in- 
dicated below. 


FIG.  70. 


§85] 


STATICS 


163 


It  is  required  to  find  analytically  the  resultant  of  the  five  forces 
given  in  §  84.     The  work  is  put  in  tabular  form: 


Magnitude 

Angle 

Cosine 

Sine 

Jf-component 

F-component 

Pi 

3 

0° 

1.000 

0.000 

3.000 

0.000 

P2 

6 

30° 

0.866 

0.500 

5.196 

3.000 

P3 

2 

70° 

0.342 

0.940 

0.684 

1.880 

p< 

2 

210° 

-  0.866 

-  0.500 

-  1.732 

-  1.000 

Pi 

4 

170° 

-  0.985 

0.174 

-3.940 

0.696 

Sum 

3.208 

4  576 

In  the  first  column  are  placed  the  letters  indicating  the  forces. 
In  columns  two  and  three  are  placed,  respectively,  the  magnitudes 
and  angles  of  these  forces.  In  the  fourth  and  fifth  columns  are 
placed,  respectively,  the  cosines  and  sines  of  these  angles.  In 
the  sixth  column  are  placed  the  products  of  the  magnitudes  of  the 
forces  by  the  cosines  of  the  angles.  In  the  seventh  column  are 
placed  the  products  of  the  magnitudes  of  the  forces  by  the  sines  of 
the  angles.  In  columns  six  and  seven,  then,  are  given,  respectively, 
the  x-  and  ^-components  of  the  five  forces.  Below  are  given  the 
algebraic  sum  of  the  z-components  and  the  algebraic  sum  of  the 
^/-components.  The  five  forces  may  be  considered  as  being  re- 
placed by  two;  one  of  3.208  pounds  acting  to  the  right  along 
the  X-axis,  the  other  of  4.576  pounds  acting  up  along  the  7-axis. 
It  now  remains  to  unite  these  two  forces  into  one  resultant,  the 
resultant  of  the  five  forces.  Since  the  two  components  of  3.208 
and  4.576  pounds  are  mutually  at  right  angles,  the  parallelogram 
of  forces  (See  Fig.  70)  is  a  rectangle;  hence,  the  magnitude 
of  the  resultant  R  =  ^(3.208) 2  +  (4.576) 2  =  5.59.  The  direc- 
tion of  OC  is  given  by  tan  6  =  ^~  =  1.426,  or,  6  =  54°  58'. 

Of  the  two  values  of  the  angle  corresponding  to  tan  6,  the 
student  must  be  careful  to  select  the  correct  value  for  6,  by  noting 
the  quadrant  within  which  the  resultant  lies. 

Exercises 

1.  Solve,  analytically,  placing  work  in  tabular  form,  the  exercises 
given  in  §  84. 


164 


MATHEMATICS 


86.  The  Simple  Crane.  To  illustrate  the  application  of  the 
polygon  of  forces  to  mechanical  problems,  let  us  consider  a  simple 
crane.  AB,  Fig.  71,  is  the  post,  CE  is  the  jib,  and  DE  the  rope 
attached  to  the  jib  at  the  point  E.  At  the  end  of  the  jib  is  sus- 
pended a  weight  of  10  tons.  It  is  required  to  find  the  tension  in 
the  rope  and  the  compression  in  the  jib.  The  point  E  is  acted  upon 
by  three  forces:  one,  of  magnitude 
10,  acts  down;  one,  of  unknown  JD 
magnitude,  acts  in  the  direction 
CE,  and  the  third,  of  unknown 
magnitude,  acts  in  the  direction 
ED.  Suppose  CD  =  12  feet, 


\OT 


FIG.  71. 


FIG.  72. 


CE  =  8  feet,  and  DE  =  7  feet.  Draw  a  triangle  to  scale  with 
sides  proportional  to  12, 8,  and  7,  Fig.  72.  Draw  the  side  propor- 
tional to  12  vertical.  Draw  the  side  proportional  to  7  through  the 
upper  end,  and  the  side  proportional  to  8  through  the  lower  end, 
of  the  vertical  side.  The  triangle  CED  is  drawn  to  scale.  A  vec- 
tor representing  the  10- ton  force  will  point  in  the  direction  DC; 
a  vector  representing  the  compression  (call  it  PI)  in  the  jib  will 


STATICS  165 

point  in  the  direction  CE,  and  a  vector  representing  the  tension 
(call  it  P2)  in  the  rope  will  point  in  the  direction  ED. 

Draw  the  polygon  of  forces,  DC'E',  for  the  three  forces,  10, 
PI,  and  P2.  This  polygon  must  close  if  the  point  E  is  in  equi- 
librium. Scale  off  the  length  C'E',  and  the  length  E'D,  which 
give,  respectively,  the  magnitude  of  the  compression  in  the  jib  and 
of  the  tension  in  the  rope.  If  DC'  is  made  10  inches  long,  or  1 
inch  representing  1  ton,  C'E'  is  found  to  be  approximately  6.67 
inches  and  DE'  approximately  5.83  inches.  This  gives  a  compres- 
sion in  the  jib  of  6.67  tons,  and  a  tension  in  the  rope  of  5.83  tons. 

Exercises 

Find  graphically  the  compression  in  the  jib,  and  the  tension  in  the 
rope,  Fig.  71,  if  CD  =  12  feet,  CE  =  8  feet,  and  ED  =  (a)  8  feet,  (6) 
9  feet,  (c)  10  feet,  (d)  11  feet,  (e)  12  feet,  (/)  13  feet,  and  (g)  14  feet. 
Plot  two  curves  upon  a  sheet  of  rectangular  coordinate  paper,  one 
representing  the  compression  in  the  jib,  the  other  representing  the 
tension  in  the  rope.  Plot  forces  as  ordinates,  and  lengths  of  DE  as 
abscissas.1 

87.  Parallel  Forces;  Moment  of  Force.  A  force  applied  to  a 
body  may  act  throughout  the  whole,  or  a  part,  of  its  mass;  or  it 
may  act  over  the  whole,  or  a  part,  of  its  surface.  A  book  lies 
upon  the  table.  Gravitational  force  acts  throughout  its  entire 
mass,  whereas  the  force  exerted  upon  it  by  the  table  acts  only  over 
that  portion  of  its  surface  in  contact  with  the  table.  If  the  sur- 
face over  which  the  force  acts  is  small  compared  with  other  dimen- 
sions, the  force  is  treated  as  though  it  acted  at  a  single  point  of  the 
body.  Thus,  in  Fig.  71,  the  post  of  the  crane  presses  to  the  right, 

1  This  problem  may  be  solved  more  easily  by  making  uae  of  the  theorem  in  ge- 
ometry which  says  that  homologous  sides  of  similar  triangles  are  in  proportion.     In 
Fig.  72,  the  triangles  DCE  and  DC'E'  are  similar,  and 
DC'       C'E'       DE' 

DC  "  CE  "  DE' 

or 

10       C'E'       DE' 

12  *  ~F  "    ~7~' 
or 

C'E'  -  6J 
DE'  -  5.833 
Solve  the  above  exercises  by  this  method. 


166  MATHEMATICS 

against  the  ceiling,  and  the  ceiling  presses  against  the  crane  with  a 
force  of  equal  magnitude,  but  in  the  opposite  direction.  This 
force,  PS,  is  distributed,  uniformly  or  non-uniformly,  over  the 
entire  surface  of  the  post  in  contact  with  the  ceiling;  but  this 
bearing  surface  is  so  small  compared  with  other  dimensions  of  the 
crane  that  the  force  P$  is  considered  acting,  and  is  said  to  act,  at  a 
point. 

A  weight  of  10  tons  is  suspended  from  the  end  of  the  jib.  Its 
line  of  action  is  vertical  and  passes  through  the  point  E.  If 
the  cable  suspending  the  weight  were  two,  three,  or  any  number 
of  times  as  long,  the  weight  would  still  exert  the  same  influence 
upon  the  crane.  For  the  cable  still  pulls  upon  the  crane  with  a 
force  of  10  tons  and  in  a  downward  direction.  The  line  of  action 
remains  the  same,  i.e.,  vertical  and  through  the  point  E.  In 
considering  forces  acting  upon  a  body,  the  position  of  the  line  of 
action,  and  not  the  exact  point  of  application  of  the  force,  is 
what  enters  into  the  solution  of  the  problem. 

It  is  readily  seen  that  the  line  of  action  of  the  resultant  of  two  forces 
passes  through  the  point  of  intersection  of  their  lines  of  action. 

In  finding  the  resultant  of  two  parallel  forces  it  becomes  neces- 
sary to  consider  two  cases:  when  the  forces  act  in  the  same  direc- 
tion and  when  the  forces  act  in  opposite  directions. 

88.  Two  Parallel  Forces  Acting  in  the  Same  Direction.  Let 
AP,  Fig.  73,  be  a  vector  representing  a  force,  P,  whose  line  of 
action  passes  through  the  point  A.  Let  BQ  be  a  vector  represent- 
ing a  force,  Q,  whose  line  of  action  passes  through  the  point  B. 
P  and  Q  are  parallel  forces  acting  upon  a  body  in  the  same  direc- 
tion. The  resultant  of  P  and  Q  together  with  its  line  of  action  is 
to  be  found.  At  the  points  A  and  B  insert  two  forces,  S  and  S', 
having  the  same  line  of  action,  equal  in  magnitude  but  acting  in 
opposite  directions.  The  resultant  of  these  two  forces  is  zero, 
hence  their  introduction  in  no  way  changes  the  resultant  force 
acting  upon  the  body.  If  AS  and  BS'  represent  the  forces  S  and 
iS',  AP'  represents  P',  the  resultant  of  P  and  S;  and  BQ'  represents 
Q',  the  resultant  of  Q  and  S'.  Since  S  and  S'  balance,  the  result- 
ant of  P'  and  Q'  is  the  resultant  of  P  and  Q.  The  lines  of  ac- 
tion of  P'  and  Q'  intersect  at  C.  At  the  point  C  resolve  the 
force  P'  (represented  by  CP")  into  its  original  components  P 


STATICS 


167 


and  S,  represented  by  CP"'  and  CS",  respectively.  At  the  point 
C  resolve  Q'  (represented  by  CQ")  into  its  original  components  Q 
and  Sf,  represented  by  CQ'"  and  CS'",  respectively.  The  forces 
S  and  S'  again  balance,  leaving  the  resultant  of  P  and  Q  equal  to 
the  vector  sum  of  CQ'"  and  CP'".  From  similar  triangles  it  is 
seen  that  CQ'"  is  equal  and  parallel  to  BQ,  and  that  CP'"  is 


-V-S' 


equal  and  parallel  to  AP.     Hence,  the  resultant  of  P  and  Q  is 
P  +  Q,  whose  line  of  action  passes  through  the  point  C  and  is  par- 
allel to  the  lines  of  action  of  P  and  Q. 
From  similar  triangles: 

AP  _  CK          BQ  _  CK 
AS  ~  AK  a       BS'  ~  BK' 

From  these  two  equations,  and  since  AS  =  BS',  it  follows  that 


_ 
BQ  ~  AK' 

If  a  is  the  distance  of  K  from  the  line  of  action  of  P,  and  if  6  is 


168  MATHEMATICS  [§88 

the  distance  of  K  from  the  line  of  action  of  Q,  it  follows  from  similar 
triangles  that 

b       BK 

a~  AK' 

Uniting  this  equation  with  the  next  preceding, 

AP  _b 

BQ  ~  a' 
or 

P  _  b 
Q  ~  a' 
or 

Pa  =  Qb. 

The  product  of  the  magnitude  of  a  force  by  the  distance  of  its 
line  of  action  from  a  fixed  point  is  called  the  moment  of  the 
force  about  the  point.  The  point  about  which  moments  are  taken 
is  called  the  origin  of  moments,  and  the  perpendicular  distance  from 
the  origin  to  the  line  of  action  of  the  force  is  called  the  arm  of  the 
force. 

If  the  relation  of  the  direction  of  the  force  to  the  origin  of 
moments  is  such  as  to  suggest  positive  rotation  about  the  origin,  the 


FIG.  74. 

moment  is  considered  positive.  If  the  relation  is  such  as  to  sug- 
gest negative  rotation,  the  moment  is  considered  negative. 
Thus,  the  moment  of  the  force  P  about  the  point  0,  Fig.  74,  is 
positive;  the  moment  of  the  force  Q  about  0  is  negative. 

The  moment  of  the  force  Q  about  the  point  K,  Fig.  73,  is  Q  6; 
the  moment  of  the  force  P  about  the  same  point  is  —  P  a.  The 
(algebraic)  sum  of  the  moments  is  Q  b  —  P  a,  but  since  in  magni- 
tude P  a  =  Q  b,  this  sum  is  zero.  In  other  words,  the  sum  of  the 


STATICS 


169 


moments  of  two  parallel  forces  acting  in  the  same  direction  about 
any  point  in  the  line  of  action  of  their  resultant  is  zero. 

Let  0,  Fig.  75,  be  any  point  in  the  plane.  Let  c  be  its  perpen- 
dicular distance  from  the  line  of  action  of  R,  the  resultant  of  the 
parallel  forces  P  and  Q;  and  d  its  perpendicular  distance  from  the 


FIG.  75. 

line  of  action  of  the  force  Q;  and  e  its  perpendicular  distance  from 
the  line  of  action  of  the  force  P.     Then 


and  since  R  =  P  +  Q, 
and  since  Pa  =  Qb, 


c  =  e  —  a 
Re  =  Re  —  Ra 

Re  =  Pe  +  Qe-  Pa  -  Qa 

Re  =  Pe  +  Qe  -  Qb  -  Qa 
Re  =  Pe  +  Q(e  -  b  -  a) 


and  since  e  —  b  —  a  =  d, 

Re  =  Pe  +  Qd. 

This  equation  shows  that  the  moment,  about  any  point,  of  the 
resultant  of  two  parallel  forces  acting  in  the  same  direction  is  equal 
to  the  sum  of  the  moments  of  the  two  forces  about  the  same  point. 
The  proof  given  above  considers  the  lines  of  action  of  the  two  given 
forces,  and  the  line  of  action  of  their  resultant,  to  be  upon  one  side 
of  the  point  0.  The  student  may  easily  show  that  the  same  law 
holds  when  the  point  0  is  taken  between  the  lines  of  action  of  the 
two  given  forces,  if  by  sum  is  meant  algebraic  sum. 


170 


MATHEMATICS 


[§89 


89.  Two  Parallel  Forces  Acting  in  Opposite  Directions.  Let 
P  and  Q,  Fig.  76,  be  two  parallel  forces  acting  in  opposite  direc- 
tions. By  analysis  similar  to  that  of  the  preceding  section,  the 
student  will  show:  that  the  resultant  of  the  two  forces  is  the  alge- 
braic sum  of  P  and  Q;  that  its  line  of  action  is  parallel  to  that  of  P 
and  Q,  and  so  placed  that  the  moment  of  the  resultant  about  any 
point  is  the  algebraic  sum  of  the  moments  of  P  and  Q  about  the 
same  point. 

The  above  proof  fails  when  P  and  Q  are  equal  in  magnitude,  for 
then  AP'  and  BQ'  are  parallel  and  the  point  C  is  at  infinity.  In 


FIG.  76. 

this  case  the  algebraic  sum  of  the  two  forces  is  zero,  and  they  can- 
not be  replaced  by  a  single  force  whose  effect  upon  the  body  is 
equivalent  to  that  of  the  two  forces. 

A  system  of  two  non-collinear  parallel  forces,  equal  in  magnitude, 
and  acting  in  opposite  directions,  is  called  a  couple.  The  product 
of  the  magnitude  of  the  forces  by  the  perpendicular  distance 
between  their  lines  of  action  is  called  the  moment  of  the  couple.  If 
the  couple  tends  to  produce  positive  rotation  its  moment  is  posi- 
tive; if  negative  rotation,  its  moment  is  negative. 


§90]  STATICS  171 

The  student  will  show  that  the  moment  of  a  couple  is  equal  to 
the  algebraic  sum  of  the  moments  of  the  two  forces  of  the  couple 
about  any  point.  He  will  also  show  that  the  algebraic  sum  of  the 
moments  of  two  equal  parallel  forces  acting  in  opposite  directions 
is  the  same  for  all  origins  of  the  moments. 

90.  A  System  of  Parallel  Forces  is  Reducible  Either  to  a  Single 
Force  or  to  a  Single  Couple.  In  a  system  of  three  or  more  parallel 
forces  there  must  be  at  least  two  that  do  not  form  a  couple. 
These  two  forces  may  be  replaced  by  their  resultant  force,  as  was 
shown  in  §§  88  and  89.  The  magnitude  of  the  resultant  is  the 
algebraic  sum  of  the  two  components  and  its  moment  about  any 
point  is  the  algebraic  sum  of  the  moments  of  the  two  forces  about 
the  same  point.  In  this  way  the  system  of  forces  is  replaced  by 
another  system  having  a  number  of  forces  less  by  one.  The 
algebraic  sum  of  the  forces  and  the  algebraic  sum  of  thier 
moments  about  any  point  remain,  however,  the  same. 

By  continuing  this  process  of  reducing  the  number  of  forces, 
the  system  is  eventually  replaced  by  one  consisting  of  only  two 
parallel  forces.  This  system,  if  not  a  couple,  can  be  replaced  by  a 
single  force,  the  resultant  of  the  two  forces.  Thus,  the  resultant 
of  a  system  of  parallel  forces  is  either  a  couple  or  a  single  force. 

The  resultant  can  be  a  single  couple  only  when  the  algebraic 
sum  of  the  given  forces  is  zero,  and  when  the  algebraic  sum  of  the 
moments  of  all  the  given  forces  about  any  point  is  different  from 
zero.  For,  by  the  above  method  of  reduction  of  the  number  of 
forces  of  the  system  one  at  a  tune,  the  algebraic  sum  of  the  magni- 
tudes of  the  forces  remains  unchanged,  and  since  this  sum  is  zero 
when  a  couple  is  obtained  it  must  have  been  zero  at  the  beginning; 
the  algebraic  sum  of  the  moments  about  any  point  remains  un- 
changed, and  since  the  moment  of  the  couple  cannot  be  zero,  the 
sum  of  the  moments  of  the  given  forces  cannot  be  zero. 

91.  System  of  Parallel  Forces  in  Equilibrium.  A  system  of 
forces  applied  to  a  body  are  said  to  be  in  equilibrium  if  the  state  of 
motion  of  the  body  is  unaltered  by  the  application  of  the  forces. 

If  the  algebraic  sum  of  a  system  of  parallel  forces  is  not  zero,  the 
resultant  of  the  system  is  a  single  force  and  the  system  is  not  in 
equilibrium.  One  of  the  necessary  conditions  for  equilibrium  of  a 
system  of  parallel  forces  is,  then,  that  the  algebraic  sum  of  the 


172 


MATHEMATICS 


[§91 


forces  be  zero.  If  the  algebraic  sum  of  the  moments  of  the  forces 
about  any  point  is  not  zero  while  the  algebraic  sum  of  the  forces  is 
zero,  the  resultant  is  a  couple  and  the  body  is  not  in  equilibrium. 

The  necessary  and  sufficient  conditions  for  a  system  of  parallel 
forces  to  be  in  equilibrium  are:  The  algebraic  sum  of  the  forces  must 
be  zero;  and  the  algebraic  sum  of  the  moments  of  the  forces  about  any 
point  must  be  zero. 

The  above  law,  it  will  be  noticed,  includes,  as  a  special  case,  the 
law  of  the  lever. 

Exercises 

1.  A  lever  14  feet  long  is  used  for  raising  a  weight  of  500  pounds. 
Where  may  the  fulcrum  be  placed  if  a  downward  pressure  of  100 
pounds  is  available  at  the  other  end  of  the  lever? 

2.  The  two  end  clevis  holes  of  a  double-tree  are  14  inches  and  20 
inches  from  the  middle  clevis  hole.     The  team  is  pulling  300  pounds. 
How  much  does  each  horse  pull? 

3.  What  must  be  the  ratio  of  the  longer  arm  to  the  shorter  arm  of  a 
double-tree:  (a)  if  one  horse  is  to  pull  5  percent  more  than  the  other 

horse ;  (6)  if  one  horse  is  to  pull 
5  percent  less  than  the  other 
horse?  If  the  team  pulls  300 
pounds,  how  much  does  each 
horse  pull  in  each  of  the  above 
cases  ? 

4.  A  horizontal  bar,  AB,  30 
B  feet  long  weighs  30  pounds  per 
linear  foot.  At  the  point  A  is 
suspended  a  weight  of  10  tons, 
and  at  the  point  B  is  suspended 
a  weight  of  3 5  tons.  Find  the 
magnitude,  the  direction,  and 
the  line  of  action  of  the  resultant  of  the  two  weights  and  the  gravi- 
tational forces  acting  upon  the  bar. 

HINT:  The  line  of  action  of  the  gravitational  forces  acting  upon  the 
particles  of  the  bar  passes  through  the  geometric  center  of  the  bar. 

5.  Let  AB,  Fig.  77,  be  a  plank  hinged  at  the  point  A,  and  held  in  a 
horizontal  position  by  a  vertical  rope.  The  plank  is  20  feet  long  and 
weighs  110  pounds.  The  rope  is  fastened  to  the  plank  15  feet  from  the 
hinge.  Find  the  tension  in  the  rope  when  a  man  weighing  150  pounds 
stands  upon  the  plank  x  feet  from  the  hinge. 


FIG.  77. 


§91]  STATICS  173 

HINT:  If  PI  represents  the  tension  in  the  rope,  taking  mo- 
ments about  A  as  origin  gives  15Pi  —  (110)(10)  —  150x  =  0,  or 
PI  =  10.r  +  220/3.  For,  the  system  of  parallel  forces  is  in  equilibrium, 
and  the  algebraic  sum  of  the  moments  about  any  point  must  be  zero. 
The  above  equation,  which  is  linear,  expresses  the  relation  between 
the  tension  in  the  rope  and  the  distance  of  the  man  from  the  point  A. 
When  the  man  is  at  the  point  A,  the  tension  is  73J  pounds.  When  he 
is  at  the  point  B,  the  tension  is  273  J  pounds.  P2  +  PI  —  HO  —  150  = 
0,  or  P2  =  360  -  Pi,  or  P2  =  286§  -  10z,  which  gives  the  reaction 
at  the  hinge.  For,  by  §  90  the  sum  of  the  parallel  forces  must 
be  zero.  When  the  man  is  at  the  point  B,  P2  =  865  pounds. 


200^ 

FIG.  78. 

6.  A  weight  of  200  pounds  is  to  be  raised  by  a  lever  x  feet  long, 
Fig.  78.  The  distance  from  the  weight  to  the  fulcrum  is  3  feet  and  the 
lever  weighs  4  pounds  to  the  foot.  If  the  lever  is  very  short  there  is 
but  little  advantage  gained  by  its  use,  whereas  if  it  is  very  long  its 
increased  weight  will  more  than  compensate  the  gain  in  leverage. 
Evidently  there  is  some  length  which  will  make  the  force  P  a  minimum. 

2(300  +  x2) 

Show  that  P  = — .     Find  the  values  of  P  when  x  equals 

x 

6,  7,  8,  9,  10,  11,  12,  13,  14,  15,  16,  17,  17£,  18,  19,  and  20.     Plot 
results  upon  a  sheet  of  rectangular  coordinate  paper. 

HINT:     When  substituting  in  the  formula,  place  it  in  the   form 


»(?+•) 


7.  The  end  holes  A  and  B  of  a  double-tree,  Fig.  79,  are  2a  inches 
apart.  The  center  hole  C  is  placed  d  inches  in  front  of  the  mid-point 
of  AB.  If  one  horse  is  pulling  ahead  of  the  other  as  indicated  in  Fig. 
79,  find  what  percent  more  the  rear  horse  is  pulling  than  the  forward 
horse. 
HINT: 

Pi  (a  cos  a  —  d  sin  a)  =  P2(a  cos  a  +  d  sin  a) 

100(Pi  -  P2)  _        20Qd  sin  « 

Pj  a  cos  a  —  d  sin  at 


174 


MATHEMATICS 


[§92 


8.  Same  as  exercise  7,  but  with  a  =  13  inches,  d  =  3  inches,  and 
=  30°. 

9.  With  the  statement  of  exercise  7,  the  difference  between  PI  and 
z  is  what  percent  of  the  load? 


FIG.  80. 

92.  The  Sum  of  the  Moments  of  Two  Non-parallel  Forces. 

Let  AP  and  AQ,  Fig.  80,  represent  two  forces.    Let  AR  represent 
their  resultant  R.    The  moment  of  P  about  0  is  the  magnitude  of 


§92]  STATICS  175 

P,  (AP),  multiplied  by  the  perpendicular  distance  from  0  to  the  line 
of  action  of  P.  This  moment  is  then  represented  by  twice  the 
area  of  the  triangle  OAP.  Similarly,  the  moment  of  Q  about  0  is 
represented  by  twice  the  area  of  the  triangle  OAQ;  and  the  mo- 
ment of  R  by  twice  the  area  of  the  triangle  OAR.  Draw  PP',  RR', 
and  QQ'  perpendicular  to  AO;  and  QB  perpendicular  to  RR'. 

The  area  of  the  triangle  OAP  =  %(PP'  -AO). 

The  area  of  the  triangle  OAQ  =  %(QQ'  -AO). 

The  area  of  the  triangle  OAR  =  %(RR'  -AO). 

RR   =  RB  -j-  BR  , 
or 

RR'  =  PP'  +  QQ'. 
By  uniting  these  equations 
2  (area  of  A  OAP)  +  2  (area  of  A  OAQ)  -  2  (area  of  A  OAR)  = 

2(PP'  +  QQ'  -  RR')  =  0, 

which  shows  that  the  moment  of  the  resultant,  R,  is  equal  to  the 
sum  of  the  moments  of  P  and  Q. 

In  the  above  proof,  the  point  0  was  so  located  that  the  algebraic 
signs  of  the  three  moments  were  the  same.  The  student  will 
prove  the  formula  for  the  case  in  which  0  is  taken  between  the 
sides  of  the  angle  QAP.  When  this  has  been  done  it  will  be  seen 
that  "sum"  in  the  above  law  means  algebraic  sum. 

From  the  above  theorem  and  from  what  has  been  proved  regard- 
ing moments  of  parallel  forces,  it  will  be  seen  that  the  algebraic 
sum  of  the  moments  of  any  number  of  forces  is  equal  to  the  moment  of 
their  resultant,  and  the  algebraic  sum  of  the  moments  of  any  number 
of  forces  in  equilibrium  is  zero. 

A  restatement  of  the  above  conclusions  follows:  The  resultant 
of  a  system  of  coplanar  forces  (not  in  equilibrium)  is  either  a  single 
force  or  a  couple.  If  the  resultant  is  a  single  force,  its  magnitude 
and  direction  may  be  found  graphically  by  finding  the  sum  of  the 
vectors  representing  the  given  forces;  its  line  of  action  is  such  that 
its  moment  about  any  point  is  the  algebraic  sum  of  the  moments  of 
the  several  forces  about  the  same  point.  If  the  resultant  is  a 
couple,  the  vector  sum  of  the  vectors  representing  the  forces 
is  zero,  and  the  algebraic  sum  of  the  moments  of  the  several 
forces  about  any  point  is  equal  to  the  moment  of  the  couple.  For 


176  MATHEMATICS  [§93 

a  system  of  coplanar  forces  to  be  in  equilibrium,  the  vector  sum  of 
the  vectors  representing  the  several  forces  must  be  zero,  and  the  alge- 
braic sum  of  the  moments  of  the  several  forces  about  any  point  must 
be  zero.  The  first  condition  throws  out  the  possibility  of  the 
resultant  being  a  single  force.  The  second  condition  throws 
out  the  possibility  of  the  resultant  being  a  couple. 

93.  Equations  of  Equilibrium.  In  the  preceding  sections  it  was 
shown  that  if  a  system  of  coplanar  forces  is  in  equilibrium,  the 
algebraic  sum  of  the  moments  of  the  forces  about  any  point  must  be 
zero.  This  law  is  generally  written  in  the  form  of  an  equation: 

2Pa  =  0 

P  represents  the  magnitude  of  force,  a  its  moment  arm,  and  2 
(the  Greek  letter  Sigma)  is  a  symbol  meaning  to  take  the  alge- 
braic sum.  Another  condition  for  equilibrium  is  that  the  vector 
sum  of  the  vectors  representing  the  given  forces  must  be  zero. 
Another  way  of  stating  this  law  is  that  if  the  given  forces  were  con- 
sidered concurrent,  their  resultant  would  be  zero.  It  was  shown, 
however  (§  85),  that  one  method  of  finding  the  resultant  of  a 
system  of  concurrent  coplanar  forces  is  first  to  resolve  each  force 
into  its  x-  and  ^-components,  and  from  the  algebraic  sum  (X)  of 
the  z-components  and  the  algebraic  sum  (F)  of  the  ^/-components 
compute  the  magnitude  of  the  resultant  (R)  by  the  formula 
R  =  \''X2  +  F2.  If  R  is  zero,  X  and  Y  separately  equal  zero. 
This  law  is  generally  written  in  the  form  of  two  equations: 

2X  =  0 
and 

27  =  0 

The  symbol  2  (the  Greek  letter  Sigma)  means  to  take  the  alge- 
braic sum.     X  in  the  above  equation  means  z-component.     2Z 
then  means  the  algebraic  sum  of  the  z-components.     2F  means 
the  algebraic  sum  of  the  y-components. 
The  equations 

2Z  =  0 
2F  =  0 
and 

2Pa  =  0 


§93] 


STATICS 


177 


are  called  the  equations  of  equilibrium  for  a  system  of  coplanar 
forces. 

The  application  of  the  equilibrium  equations  will  be  illustrated 
by  examples: 

EXAMPLE  1:  The  crane,  Fig.  71,  considered  as  a  whole,  is  acted 
upon  by  four  forces:  viz.,  the  downward  force  of  10  tons,  the 
upward  push  of  the  floor,  P^,  the  push  of  the  floor  to  the  right, 
PS;  and  the  push  of  the  ceiling  to  the  left,  P3.  P4,  Ps,  and  PS  are 
unknown  in  magnitude,  and  are  to  be  found.  Let  AB  =  12  feet 
and  FE  =  4  feet.  Then 

2Z  =  P5  -  P3  =  0 
ZF  =P4  -  10  =  0 

and  by  taking  moments  about  the  point  A, 

ZPa  =  12P3  -  10-4  =  0, 

or  PS  =  10/3.     The  forces  Pt  and  P5  contribute  nothing  to  the 

moment  equation,  for  they  act 

through  the  point  A  about  which 

moments  are  taken;  hence  their 

moment  arms  and  moments  are 

zero. 

The  second  equation  gives  P* 
=  10  tons. 

The  first  equation  gives  P3  = 
PS  =  10/3  tons.  This  means 
that  the  post  of  the  crane 
pushes  to  the  left  against  the 
flange  of  the  floor  with  a  force 
of  31  tons.  The  construction  at 
this  point,  as  well  as  at  the  ceil- 
ing, must  be  such  as  to  with- 
stand this  pressure. 

EXAMPLE.  2:  Let  AB,  Fig.  81, 
be  a  ladder  16  feet  long,  resting 
against  a  smooth  vertical  wall, 
CB.  The  lower  end  of  the 

ladder  rests  upon  the  horizontal  ground,  6  feet  from  the  building. 
CB  is  14.83  feet  long.     The  ladder  weighs  35  pounds,  and  a  man 

12 


FIG.  81. 


178 


MATHEMATICS 


[§93 


who  ascends  weighs  150  pounds.     Find  the  pressure  of  the  upper 
end  of  the  ladder  against  the  wall  when  the  man  stands  on  a  rung 
5  feet  from  the  point  A. 
By  taking  moments  about  the  point  A, 

SPa  =  +35-3  +  150-15/8  -  14.83  Pi  =  0, 
or  PI  =  26.04  pounds.     (The  weight  of  the  ladder  is  considered 
as  a  single  force  acting  through  the  middle  point.) 


FIG.  82. 

Taking  moments  about  the  point  A,  letting  x  be  the  distance 
(measured  along  AB)  the  man  stands  from  the  point  A, 

SPa  =  35-3  +  150?|  -  14.83  Pi  =  0, 


or 


Pi  = 


_  f225- 


L  4 


x  +  105    -s-  14.83. 


From  this  equation  PI  can  be  calculated  for  all  values  of  x, 
i.e.,  for  any  position  of  the  man  upon  the  ladder.  The  equation 
is  linear  and  hence  its  graphic  representation  upon  a  sheet  of 
squared  paper  is  a  straight  line.  In  order  to  draw  the  straight 
line  all  that  is  needed  are  the  coordinates  of  two  points.  These 
may  be  found  by  giving  to  x  two  values,  as  0  and  16,  and  calcu- 
lating the  corresponding  values  of  PI. 

EXAMPLE  3 :  Let  all  members  of  the  bridge  truss,  Fig.  82,  be 
equal  in  length,  k  feet.  The  bridge  sustains  loads  as  indicated. 
Find  the  reactions,  PI  and  P2,  at  the  points  A  and  B.  By  taking 
moments  about  the  point  A,  it  follows  that 


§94]  STATICS  179 

k  3k  *>k 

3k  P2  -  1000  2  -  4000  k  -  1000  y  -  5000  2fc  -  1000  ^ 

-  3000  -3k  =  0, 
or 

3P2  -  500  -  4000  -  1500  -  10,000  -  2500  -  9000  =  0, 
or 

3P2-  27,500  =  0, 
or 

P2  =  9166f  pounds. 
SF  =  Pi  +  P2  -  2000  -  1000  -  4000  -  1000  -  5000  -  1000  - 

3000  =  0, 
or 

P!  +  P2  =  17,000, 
or 

Pi  +  9166|  =  17,000, 
or 

PX  =  7833i 

Exercises 

1.  Find  the  magnitudes  of  the  forces  PI,  P2,  and  Pa,  Fig.  83,  neglect- 
ing the  weight  of  the  crane. 

2.  Find  the  magnitudes  of  the  forces  PI  and  P2,  Fig.  84,  also  the 
tension  in  the  rope. 

HINT:  Draw  ABC  to  scale,  using  1/2  inch  to  represent  a  foot. 
Neglect  the  weight  of  AC. 

[94.]  Tension  and  Compression.  The  tension  or  compression  in 
members  of  a  construction  may  be  easily  computed,  if  the  weights 
of  the  members  are  neglected,  if  the  loads  are  applied  at  the  joints, 
and  if  the  members  are  discontinuous  at  the  joints.  By  discon- 
tinuous is  meant  that  a  single  piece  does  not  form  a  side  of  more 
than  one  polygon.  Thus,  in  the  roof  truss,  Fig.  85,  the  tie  AB  con- 
sists of  three  pieces,  AD,  DE,  and  EB,  bolted  together  at  the  points 
D  and  E. 

If  the  loads  are  applied  at  the  joints,  the  tension  or  compression 
in  every  member  is  the  same  throughout  the  entire  length  of  that 
member,  and  the  line  of  action  of  the  compressive  or  tensile  force 
within  the  member  is  that  of  the  longitudinal  dimension  of  that 
member. 


180 


MATHEMATICS 


[§94 


The  method  of  finding  tension  and  compression  will  be  illus- 
trated by  solving  the  problem  represented  by  Fig.  85. 
Let  AD  =  DE  =  EB  =  lOf  feet 


/\  10  Tons 


Let 
and 


FIG.  83. 

CK  =  8  feet 

AC  =  BC  =  17.88  feet 

CD  =  CE  =  9.61  feet 


§94] 


STATICS 


181 


To  find  the  reactions,  PI  and  Pt,  of  the  supports  use  2  Y  =  0, 
and      2Po  =  0,  taking  moments  about  the  point  A. 


FIG.  84. 


/\  10  Tons 


or 


FIG.  85. 

27  =  Pi  +  P2  -  500  -  2000  -  1000  -  1000  -  500  =  0, 
P!  +  P2  =  5000. 


182  MATHEMATICS  [§94 

SPa  =  2000  lOf  +  1000  16  +  1000-2H  +  500-32  -  P2-32  =  0, 
or 

P2-32  =  74,667, 
or 

P2  =  2333  pounds. 
Then  PI  =  2667  pounds. 

Imagine  the  members  AC  and  AD  cut  by  the  section  a  and  the 
right-hand  portion  removed.  In  order  to  keep  the  left-hand  por- 
tion in  equilibrium,  Fig.  86,  a,  forces  P3  andP  4,  equal  in  magnitude 
to  the  tension  or  compression  in  AC  and  AD,  must  be  introduced 

c' 


A' 


FIG.  86. 

upon  the  cut  ends.  To  express  this  in  another  way:  suppose  an 
imaginary  plane  through  the  rafter  AC  divide  it  into  an  upper  and 
a  lower  part.  The  part  below  pushes  against  the  part  above,  and 
the  part  above  pushes  against  the  part  below;  the  two  compressive 
forces  are  equal  in  magnitude  and  this  magnitude  is  called  the 
compression  within  the  rafter.  If  the  upper  part  of  the  member 
were  removed,  and  if  a  force  equal  to  the  compression  were 
introduced  acting  upon  the  remaining  cut  end  oi  AC  and  in  a  direc- 
tion toward  A,  the  lower  part  of  the  rafter  would  be  acted  upon  by 
the  same  force  as  before  the  upper  portion  was  removed. 

The  four  forces  acting  upon  the  remaining  portion  of  the  truss, 
Fig.  86,  a,  are  in  equilibrium.  The  two  vertical  forces  acting 
at  the  point  A  may  be  considered  as  one  force  acting  up,  of  magni- 
tude 2167  pounds.  Draw  the  line  C'K',  Fig.  86,  6,  to  scale, 
representing  the  upward  2167-pound  force.  C'K'  is  parallel  to 
CK,  Fig.  85.  Through  C'  draw  a  line  parallel  to  AC  and  through 
K'  draw  a  line  parallel  to  AK.  These  two  lines  intersect  at  A'. 


§94]  STATICS  183 

K'C'A'  is  then  the  force  polygon  for  the  three  forces,  2167, 
Pa  and  P\.  C'A'  is  the  vector  representing  the  force  PS,  and  A'K' 
is  the  vector  representing  the  force  Pi.  The  magnitudes  of  PS  and 
P*  may  be  scaled  off  from  this  triangle.  Since  K'C'A'  is  a  closed 
polygon,  the  arrows  run  around  the  polygon  in  the  same  sense. 
The  arrow  of  C'A'  then  points  from  C"  toward  A',  which  says 
that  the  arrow  of  PS,  Fig.  86,  a,  was  chosen  to  point  in  the  proper 
direction,  or  the  member  AC  is  in  compression.  Similarly,  the 
arrow  of  P\  was  chosen  in  the  proper  direction,  or  the  member  AD 
is  in  tension. 

The  magnitudes  of  PS  and  Pi  may  be  found  by  another 
method  which  does  not  require  drawing  the  triangle  K'C'A'  to 
scale.  The  triangles  K'C'A'  and  KG  A,  Fig.  85,  are  similar,  and 

C'A'  _  K'C't 
CA   ==   KG 

or 

C'A1      2167 

17.88  "     8   ' 

or 

C'A'  =  4842, 
PS  =  4842  pounds. 

Again : 

A'K'  _  K'C^ 
~AK  ''=  KG' 

or 

A'K^  _  2167 
16  8    ' 

or 

A'K'  =  4333, 

P4  =  4333  pounds. 

To  find  the  tension  or  compression  in  the  members  CD  and  DE, 
consider  the  portion  about  the  point  D  removed  by  the  cut  6, 
Fig.  85. 


184 


MATHEMATICS 


At  the  cut  ends  insert  the  forces  PS  and  P6,  Fig.  87,  a,  equivalent 
to  the  tension  or  compression  in  the  cut  members.  The  force 
polygon  for  the  four  forces  is  represented  by  the  triangle  K'C'D', 
Fig.  87,  b.  This  triangle  is  similar  to  triangle  KCD,  Fig.  85. 


4333 


Hence 


or 


or 


Again: 


or 


43337) 


2000^ 


FIG.  87. 


D'C'       C'K' 


DC 

D'C' 
9.61 


CK' 
2000 


D' 


,  ,2000^ 
K' 


D'C'  =  2402.5, 

P5  =  2402.5  pounds  tension. 

K'D'       2000 


KD 
M'D'  +  4333 


8    ' 
2000 


or 


M'D'  =  -  3000  pounds. 


The  negative  sign  shows  that  the  arrow  of  P&  points  in  the  wrong 
direction. 

P&  =  3000  pounds  tension 

In  a  similar  way  by  means  of  cuts  c  and  d,  Fig.  85,  the  tension  or 
compression  in  EB,  BC,  and  EC  may  be  found. 


§95] 


STATICS 
Exercises 


185 


1.  Find  the  tension  or  compression  in  the  members  BC,  BE,  EC, 
and  ED,  of  the  truss  given  in  Fig.  85. 

2.  Find  the  tension  or  compression  in  each  member  of  the  truss 
given  in  Fig.  82. 

95.  Rope  and  Pulley.      Let  PI  and  P2,  Fig.  88,  be  the  tensions 
in  a  rope  upon  opposite  sides  of  a  pulley  over  which  it  passes.     If 


FIG.  88. 

there  be  no  friction  between  the  pulley  and  the  axle,  the  line  of 
action  of  the  force  exerted  by  the  axle  upon  the  pulley  will  pass 
through  the  center  of  the  pulley.  Taking  moments  about  the 
center  of  the  wheel  (calling  a  the  radius  of  the  wheel) 

Pia-P2a  =  0,    orPi=P2 


FIG.  89. 

This  says  that  if  a  flexible  rope  passes  over  a  series  of  friction- 
less  pulleys,  the  tension  at  every  point  of  the  rope  is  the  same. 
In  Fig.  89,  the  pull  on  each  end  of  the  rope  is  the  same;  and  if  the 


186 


MATHEMATICS 


[§95 


rope  be  cut  at  any  point,  K,  a  force  equal  to  P  must  be  introduced 
upon  each  free  end  of  the  rope  in  order  to  hold  the  system  in 
equilibrium. 

Fig.  90  represents  a  system  of  pulleys  by  which  a  pull,  P,  holds 
a  weight  of  10  tons  in  equilibrium.  To  find  the  tension  in  the 
rope  passing  around  pulley  E,  imagine  the  rope  cut  by  the  section 


/\ 10  Tons 


FIG.  90. 


a,  and  forces  introduced  upon  the  cut  ends  equal  to  the  magnitude 
of  the  tension  in  the  rope.  Represent  these  forces  by  PI.  The 
forces  acting  upon  the  part  of  the  body  below  the  section  are  in 
equilibrium,  and  PI  +  PI  —  10  =  0,  or  PI  =  5.  The  tension  in 
the  rope  passing  about  pulley  E  is  5  tons. 

If  a  second  cut  6  be  made,  and  if  P2  represent  the  tension  in  the 
rope  passing  about  pulley  F,  P2  +  P2  —  P:  =  0,  or  P2  +  P2  — 
5=  0,  or  P2  =  2J.  The  tension  in  the  rope  passing  about  pul- 
ley F  is  2£  tons. 


§95] 


STATICS 


187 


If  a  third  cut  c  be  made,  and  if  P  represents  the  tension  in  the 
rope  passing  around  pulleys  G  and  H ,  P  +  P  —  P2  =  0,  or 
P  +  P  -  2f  =  0,  or  P  =  l£.  The  pull,  P,  necessary  to  hold 
the  weight  10  tons  in  equilibrium  is  lj  tons. 

Exercises 

1.  Find  the  pull  P  which  will  hold,  by  means  of  the  system  of 
pulleys  shown  in  Fig.  91,  the  weight  of  10  tons  in  equilibrium.  Find 
the  pull  on  the  hooks  A,  B,  and  C. 


10  Tons 


FIG.  91. 


2.  Find  the  pull  P  which  will  hold,  by  means  of  the  system  of  pulleys 
shown  in  Fig.  92,  the  weight  of  10  tons  in  equilibrium.     Find  the  pull 
on  the  hook  A. 

3.  The  differential  wheel  and  axle  shown  in  Fig.  93  consists  of  three 
drums  of  radii  a,  b,  and  c.     A  force  P  is  applied  to  a  rope  which  unwinds 
from  the  largest  drum.     As  the  drums  rotate  a  rope  unwinds  from  the 


188 


MATHEMATICS 


[§95 


smallest  and  winds  up  upon  the  medium  sized  drum,  lifting  a  weight  of 
10  tons.  Find  P  which  will  hold  the  weight  in  equilibrium.  As  P 
moves  down  x  feet,  how  far  does  the  weight  rise? 


y'      ~\10  Tons 

FIG.  92. 


/\  10  Tons 

FIG.  93. 


CHAPTER  VII 

PERMUTATIONS,     COMBINATIONS,    AND    THE    BINOMIAL 
EXPANSION 

96.  A  Fundamental  Principle.  From  a  recitation  room  there 
are  two  doors  leading  into  the  hall;  from  the  hall  there  are  three 
exits.  In  how  many  ways  can  a  person  leave  the  building  from 
the  room?  Let  the  doors  of  the  room  be  numbered  1  and  2,  and 
let  the  exits  from  the  hall  be  marked  A,  B,  and  C.  A  person  may 
pass  through  door  1  and  exit  A;  door  1  and  exit  B;  door  1  and 
exit  C;  door  2  and  exit  A ;  door  2  and  exit  B;  or  door  2  and  exit  C. 
After  passing  through  door  1  there  are  three  possible  paths:  one 
through  exit  A;  one  through  exit  B;  and  one  through  exit  C. 
There  are  exactly  the  same  number  of  paths  passing  through  door 
2.  In  all  there  are  2  X  3  or  6  ways  of  passing  from  the  building. 

If  there  were  ra  doors  leading  from  the  room  into  the  hall,  and 
n  exits  from  the  hall,  obviously  there  would  be  m  X  n  ways  of 
leaving  the  building. 

The  following  is  a  fundamental  principle:  //  one  thing  can  be 
done  in  m  ways,  and  if  a  second  can  be  done  in  n  ways,  the  two  things 
can  be  done  in  the  order  indicated  in  m  X  n  ways. 

It  is  obvious  that  the  order  in  which  the  two  things  are  done  does 
not  affect  the  number  of  ways  in  which  they  may  be  done.  They 
may  even  be  done  simultaneously.  To  illustrate:  suppose  a 
penny  and  a  die  are  thrown  simultaneously;  the  penny  will,  fall 
in  one  of  two  ways;  the  die  will  fall  in  one  of  six  ways;  then,  the 
two  will  fall  simultaneously  in  one  of  2  X  6,  or  12  ways.  This 
illustration  shows  that  the  fundamental  principle  applies  not  only 
to  the  number  of  ways  of  doing  two  things  but  also  to  the  number 
of  ways  two  events  may  occur. 

It  will  be  seen  at  once  that  if  one  thing  can  be  done  in  m\ 
ways,  a  second  in  m*  ways,  a  third  in  m$  ways,  and  so  on,  the  num- 
ber of  ways  they  can  be  done  together  is  m\  X  m^  X  m$  X  .  .  . 

189 


190  MATHEMATICS  [§97 

97.  Permutation  and  Combination  Defined.  Suppose  we  have 
five  cubes,  one  painted  white,  one  black,  one  red,  one  blue,  and 
one  green.  From  these  five  a  selection  of  one  is  made.  This  can 
be  done  in  five  ways.  Each  time  a  selection  of  one  is  made,  a 
selection  of  four  is  left  consisting  of  a  different  combination  of 
colors. 

A  selection  of  two  cubes  is  made.  This  may  be  done  in  ten 
ways,  as: 

white  and  red,  black  and  blue, 

white  and  black,  black  and  green, 

white  and  blue,  red  and  blue, 

white  and  green,  red  and  green, 

black  and  red,  blue  and  green. 

Each  time  a  selection  of  two  cubes  is  made,  a  selection  of  three 
cubes  remains.  Then  a  selection  of  three  cubes  from  the  five 
cubes  may  be  made  in  ten  ways.  A  selection  of  four  cubes  may  be 
made  in  five  ways,  and  a  selection  of  five  cubes  may  be  made  in  one 
way. 

If  from  n  things,  a  group  of  r  (r  ^  ri)  are  selected,  the  selection 
is  called  a  combination  of  the  n  things  taken  r  at  a  time.  The 
symbol  representing  the  number  of  combinations  that  can  be  made 
of  the  n  things  taking  r  at  a  time  is  nCT.  Thus,  from  the  previous 
illustrations,  5(7i  =  5;  £2  =  10;  5Ca  —  10;  £4  =  5,  and  5^5  =  1. 

Consider  again  the  five  colored  cubes.  Select  two  by  first  pick- 
ing out  one  and  then  a  second.  This  may  be  done  in  twenty  ways 
if  the  order  of  the  selection  is  considered  (the  selection  first  black, 
then  white,  is  considered  different  from  the  selection  first  white, 
then  black).  For,  by  the  fundamental  principle,  the  first  selec- 
tion may  be  made  in  five  ways,  and  the  second  selection  in  four 
ways;  the  total  number  of  selections  possible  is  4  X  5  =  20. 

Any  selection  of  r  (r  ^  n)  things  taken  from  n  things  which 
takes  into  consideration  the  order  of  selection  is  called  a  permu- 
tation of  n  things  taken  r  at  a  time. 

The  symbol  representing  the  number  of  permutations  of  n 
things  taken  r  at  a  time  is  nPr. 

It  is  to  be  remembered  that  with  permutations  we  consider  the 
order  of  selection,  or  the  arrangement,  of  the  objects  as  well  as  the 


PERMUTATIONS,  COMBINATIONS,  ETC.  191 

individuals  which  go  to  make  up  the  selection,  while  with  combina- 
tions we  consider  only  the  individuals  of  the  selection  without 
reference  to  the  order  of  their  selection  or  arrangement.  To 
illustrate:  If  in  a  plane  there  are  five  points,  no  three  of  which  are 
upon  one  straight  line,  the  problem  of  finding  the  number  of 
lines  determined  by  the  points  is  a  problem  in  combinations. 
For  it  is  immaterial  whether  we  think  of  the  line  drawn,  for  ex- 
ample, from  the  point  A  to  the  point  B,  or  from  the  point  B  to 
the  point  A. 

From  the  five  digits  1,2,  3,  4,  and  5,  how  many  products  may 
be  formed,  using  two  and  only  two  digits  at  a  time?  This  is  a 
problem  in  combinations,  for  example,  the  product  3  X  2  is  the 
same  as  the  product  2  X  3. 

From  the  five  digits  1,  2,  3,  4,  and  5,  how  many  two-figured 
numbers  may  be  formed?  This  is  a  problem  in  permutations; 
for  example,  the  number  32  is  different  from  the  number  23. 

Exercises 

1.  Now  many  products  can  be  formed  from  the  digits  1,  2,  3,  4,  and 
5  taking  (a)  two  and  only  two  digits  to  form  a  product;  (6)  three  and 
only  three  digits  to  form  a  product;  (c)  four  and  only  four  digits  to 
form  a  product;  and  (d)  five  digits  to  form  a  product? 

2.  How  many  lines  may  be  drawn  through  five  points  in  a  plane, 
no  three  of  which  are  upon  one  straight  line? 

3.  How  many  planes  are  determined  by  five  points  in  space,  no 
four  of  which  are  on  one  plane? 

4.  How  many  points  of  intersection  are  determined  by  five  straight 
lines  on  a  plane,  if  no  three  intersect  in  the  same  point? 

5.  How  many  numbers  can  be  formed  from  the  digits  1,  2,  3,  4,  and 
5  (a)  if  two  and  only  two  digits  are  used  in  each  number;  (6)  if  three 
and  only  three  digits  are  used  in  each  number;  (c)  if  four  and  only  four 
digits  are  used;  (d)  if  five  and  only  five  digits  are  used? 

98.  Formula  for  nPr.  Let  a,  b,  c,  d,  e,  f  be  permuted  two  at  a 
time.  The  first  letter  may  be  selected  in  six  ways,  i.e.,  the  first 
letter  selected  may  be  the  a,  the  b,  the  c,  the  d,  the  e,  or  the  /. 
When  the  first  letter  is  selected,  the  second  letter  may  be  selected 
in  five  ways  from  the  remaining  five  letters.  Then,  by  the  funda- 
mental principle,  the  number  of  ways  two  letters  may  be  selected 
from  six  letters  is  6  X  5,  or  30.  Thus,  ^  =  6  X  5  =  30. 


192  MATHEMATICS  [§98 

If  three  letters  are  placed  in  each  permutation,  the  first  can  be 
selected  in  six  ways,  the  second  in  five  (6  —  1)  ways,  and  the 
third  in  four  (6  -  2)  ways.  Thus,  ePs  =  6X5X4=  120. 

If  four  letters  are  placed  in  each  permutation,  the  first  can  be 
selected  in  six  ways,  the  second  in  five  ways,  the  third  in  four  ways, 
and  the  fourth  in  three  ways.  Thus,  ^4  =  6X5X4X3  =  360. 

The  student  will  show,  by  reasoning  similar  to  the  above,  that 
ePs  =  6X5X4X3X2  =  720,  and  ^6  =  6X5X4X3X2 
X  1  =  720. 

If  n  things  are  permuted  r  at  a  time,  the  first  selection  is  made 
from  the  n  things,  or  it  may  be  made  in  n  ways;  the  second 
selection  is  made  from  the  remaining  n  —  1  things,  or  it  may  be 
made  in  n  —  1  ways.  Then,  by  the  fundamental  principle,  the 
first  two  things  may  be  selected  in  n(n  —  1)  ways.  The  third 
selection  is  made  from  the  remaining  n  —  2  things,  or  it  may  be 
made  in  n  —  2  ways.  Thus,  the  first  three  things  may  be  selected 
in  n  (n  —  !)(«  —  2)  ways.  Continuing  in  this  way  the  first  four 
things  may  be  selected  in  n(n  —  l)(n  —  2}(n  —  3)  ways,  the 
first  five  things  in  n(n  —  \)(n  —  2)(n  —  3)(n  —  4)  ways,  the 
first  six  things  in  n(n  —  l)(n  —  2)(n  —  3)(n  —  4)(n  —  5)  ways, 
and  so  on.  The  r  things  may  be  selected  in 

n(n  -  l)(n  -  2)(n  -  3)   .    •    •  [n  -  (r  -  2)][n  -  (r  -  1)] 
ways.     Therefore 

nPr  =  n(n  —  l)(n  —  2)(n  —  3)  •    •    -to  r  factors, 
or         nPr  =  n(n  -  l)(n  -  2)(n  -  3)  •    •    •  (n  -  r  +  1)          (1) 

If  r  =  n,  formula  (1)  reduces  to 

nPn  =  n(n  —  l)(n  —  3)  •    •    •  to  n  factors,  or 
nPn  =  n(n  -  l)(n  -  2)(n  -  3)  •    •    •  3  X  2  X  1 
nPn  =    n_  (2) 

where  J^  (read  "factorialn")  means 

1X2X3X4  •    •    •   (n  -  2)(n  -  l)n,  or  the  product  of  all  the 
positive  integral  numbers  from  1  up  to  and  including  n. 

The  symbol  nPn,  the  number  of  permutations  of  n  things  taken 
all  (n)  at  a  time,  will  be  written  Pn. 
From  formula  (1), 

n-iPr-i  =  (n  -  l)(n  -  2)(w  -  3)    •    •    •    (n  -  r  +  1) 


§99]  PERMUTATIONS,  COMBINATIONS,  ETC.  193 

found  by  replacing  n  by  n  —  1,  and  r  by  r  —  1.     Then 

nPr  =  n(n_1P,_1).  (3) 

By  multiplying  both  numerator  and  denominator  of  the  right- 
hand  side  of  formula  (1)  by   n—r  ,  it  reduces  to 

,P,=  -l^-.  (4) 


n  —  r 


Exercises 

1.  Solve  exercise  5  of  the  preceding  section  by  formula  (4). 

2.  How  many  different  signals  can  be  given  with  seven   flags  of 
different  colors,  by  hoisting  them  in  a  vertical  line,  any  number  at  a 
time? 

3.  How  many  different  numbers  can  be  formed  from  the  digits 
1,  2,  3,  4,  5,  6,  if  a  digit  cannot  be  used  more  than  once  in  any  one 
number? 

4.  From   twenty  men  how  many  nines  can  be  formed,  assuming 
that  every  man  can  play  any  position?     By  nine  is  not  meant  merely 
the  aggregate  of  nine  men,  but  has  reference  as  well  to  the  positions 
played. 

5.  Same  as  exercise  4,  excepting  that  only  two  can  pitch  and  that 
only  three  can  catch. 

6.  Same  as  exercise   5,  excepting  that  the  pitchers  and  catchers 
play  no  other  positions. 

99.  Formula  for  nCr.  Within  each  combination  of  n  things 
taken  r  at  a  time  there  are  _£_  permutations.  The  number  of 
permutations  of  n  things  taken  r  at  a  time  is  equal  to  the  product 
of  the  number  of  combinations  of  n  things  taken  r  at  a  time  by  the 
number  of  permutations  within  each  combination  taking  r  at  a 
time,  or 

npr  = 
or, 


or 


1:3 


n  —  r 


194  MATHEMATICS  [§100 

Every  time  a  combination  of  r  things  is  selected  from  n  things,  a 
combination  of  n  —  r  things  remains.  Thus,  the  number  of  com- 
binations of  n  things  taken  r  at  a  time  is  equal  to  the  number  of 
combinations  of  n  things  taken  n  —  r  at  a  time.  This  fact  can  be 
shown  from  formula  (1)  by  replacing  r  by  n  —  r,  when  it  reduces  to 

nCn-r  = 

n  —  r 


n 

n— 

r 

n-n  +  r 

a  formula  identical  with  (1). 

Exercises 

1.  Solve  exercises  1,  2,  3,  and  4,  §  97,  by  formula  (1). 

2.  There  are  twenty  points  in  a   plane,  no   three   of  which  are 
upon  the   same   straight  line,  excepting   five,  which  are  upon  one 
straight  line.     How  many  lines  are  determined  by  the  points? 

100.  The  Binomial  Theorem.  The  product  of  the  n  binomials, 
(fli  +  bi)(az  +  62) (as  +  63)  •  •  •  (a»  +  bn),  is  the  sum  of  the 
partial  products  formed  by  taking,  in  as  many  ways  as  possible, 
one  and  only  one  letter  from  each  of  the  parentheses. 

When  the  binomials  are  all  equal,  the  product  becomes  (a  +  &) 
(a  -f  6)  .  .  .  to  n  factors.  By  taking  the  first  letter  from  each  of 
the  parentheses  the  partial  product  a"  is  obtained.  By  taking  a 
b  from  one  of  the  parentheses,  and  an  a  from  the  remaining  n  —  1 
parentheses,  a  partial  product,  an~lb,  is  obtained.  Several  par- 
tial products  of  this  form  may  be  obtained  by  selecting  the  b  from 
different  parentheses.  In  fact,  it  may  be  selected  in  n  ways, 
giving  n  partial  products  of  the  form  an~l  b.  By  taking  the  b  from 
two  of  the  parentheses,  and  the  a  from  the  remaining  n  —  2  par- 
entheses, a  partial  product,  an~z  b-,  is  obtained.  Several  partial 
products  of  this  form  may  be  obtained  by  selecting  the  b's  from 
different  pairs  of  parentheses.  The  number  of  partial  products 
which  may  be  formed  in  this  way  is  equal  to  the  number  of  ways  in 
which  two  b's  can  be  selected  from  n  b's,  or,  nCz.  In  the  same  way 
it  may  be  shown  that  the  number  of  partial  products  of  the  form 
an-s  £3  js  B(73-  ^0  number  of  partial  products  of  the  form  an~4,  b* 
is  nCi,  etc.  Thus, 
(o  +  b)n  =  an  +  nan-lb  +  nCzan~2b2  +  nC3an~3b3 

+nCtan-W  +  .    .    .  +bn, 


§101] 
or 


PERMUTATIONS,  COMBINATIONS,  ETC. 


195 


-(---JV^-C-lKn-Vv 


n(n  -  l)(n  -  2)(n  -  3) 


an-4b4 


bn. 


This  is  known  as  the  binomial  expansion,  or  binomial  theorem. 
The  coefficients  of  the  right-hand  side  of  the  equation  are  called 
the  binomial  coefficients. 

For  n  =  2,  (a  +  6)2  =  a2  +  2ab  +  b\ 

For  n  =  3,  (a  +  fe)3  =  a3  +  3a26  +  3a62  +  b3. 

For  n  =  4,  (a  +  6)4  =  a4  +  4a36  +  6a262  +  4a63  +  64. 

For  n  =  5,  (a  +  6)5  =  a5  +  5a46  +  10a362  +  10a263  +  5a64  +  &5. 


(2 


6 


X4 


(2)3(3x2) 


6X5X4 


6X5X4X3X2Xl 


(3x2)6> 


or 

(2  +  3x2)6  s  64 


576x2 


4320x6  +  4860a;8 


If  6  is  negative,  it  will  be  seen  that  every  alternate  term  of  the 
expansion,  beginning  with  the  second,  is  negative. 

Exercises 
Expand  the  following  : 

1.  (a  -  6)*.  4.  (x2  -  3)4. 

2.  (2  +  6)6.  5.  (x*  +  3z2)6. 

3.  (2  -  3x)6.  6.  (x*  +  x-1)6. 

101.  The  Binomial  Expansion  for  Negative  and  Fractional 
Exponents.  In  the  preceding  section  the  expansion  was  given 
for  a  binomial  with  a  positive  integral  exponent.  It  is  shown  in 
higher  mathematics  that  the  expansion  also  holds  for  fractional 


196  MATHEMATICS  [§102 

and  for  negative  exponents,  providing  that  the  numerical  value  of  b 
is  less  than  the  numerical  value  of  a. 

For  a  positive  integral  exponent  the  binomial  expansion  consists 
of  a  finite  number  of  terms,  n  +  1.  For  fractional  or  negative 
exponents  the  number  of  terms  is  infinite,  or  the  expansion  is  said 
to  be  an  infinite  series. 

ILLUSTRATION  1:    Expand  (1  +  a;)"1. 

- 


^  ^(_l)-4(a.)S  + 

or 

ILLUSTRATION  2:     Expand  (1  +  a;)*  . 

(1/2}  (1/2  —  1} 
(I  _i_  xy    =  (1)     +  (1/2) (lr  a;  4-  (1)~^  (a;)2 


2 


or 


Exercises 

Expand  the  following  to  four  terms: 

1.  (1  +  x)U.  4.  (1  -  x)-1. 

2.  (1   -  x)K.  5.  (1  +  x)~*. 

3.  (1  +  x)H.  6.  (1  -  x)~*. 

102.  Approximation  Formulas.    If  a;  is  numerically  very  small  the 
expansion 

n(w  —  1)  n(n  —  l)(n  —  2) 

(1  +  x)n  ~  1  +  nx  +  ^—^ — -  a;2  +  -          I  3  -  x3  +  •    •    • 

is  approximately  equal  to 

1  +  nx. 

For,  a;2,  a;3,  .    .    . ,  are  higher  powers  of  a  small  fraction  and  hence 


§103]         PERMUTATIONS,  COMBINATIONS,  ETC.  197 

small  compared  with  x.     If  the  symbol  =  represents  "  approxi- 
mately equal," 

(l  +  x)"  =  l  +  nx  (1) 

if  x  is  numerically  very  small.     Thus,  to  illustrate, 
(1  +  0.06)  w  =  1  +  M0.06)  =  1.02. 

The  true  value  of  this  expression  is,  to  five  decimal  places,  1.01961. 
Again: 

(1  -  0.06)  *  =  1  -  1(0.06)  =  0.98. 

If  x,  y  and  z  are  numerically  very  small,  the  student  will  show 
by  multiplication  and  division  that  the  following  are  approxi- 
mate formulas: 

(2) 

(3) 

(1  +  x)(l  +  y)(l  +  z)  =  1  +  x  +  y  +  z.  (4) 

It  is  to  be  remembered  that  the  above  formulas  hold  for  nega- 
tive values  of  x,  y,  and  z  as  well  as  for  positive  values. 
Thus,  by  formula  (2) 

(1  +  0.03)  (1  -  0.05)  =  1  +  0.03  -  0.05 
=  0.98. 

Exercises 
Find  the  approximate  value  of  each  of  the  following: 


1.03 

2.  (0.98)    .  '*  1.02' 

H  _    1.03 

3.  (0.93)    .  7-  0.97* 

4.  (l.Q3)(1.02).  8.  (1.04)  (1.06)(0.95). 

103.  The  Compound  Interest  Law.     If  a  dollars  are  loaned  at  r 
percent  per  annum,  compound  interest, 

at  the  end  of  one  year  the  amount  is          a   1  +  JQQ  ' 


198  MATHEMATICS  [§103 

f          r  "12 

at  the  end  of  two  years  the  amount  is          a   1  +  JQQ     ' 

[r  "is 
1    +    JQQ          ' 

[r  ~~\* 
1     +     TQQ 

If  the  interest  is  compounded  semi-annually  instead  of  annually  the 
amount  at  the  end  of  x  years  is 

[r  "12* 
1  +  20oJ 
if  compounded  monthly  the  amount  at  the  end  of  the  same  period 

is 

[r    "1  12* 
f  I200J 
and  if  compounded  n  times  a  year  the  amount  is 

[ 
1 


Now,  if  we  find  the  limit  of  this  expression  as  n  increases  indefi- 
nitely, we  find  the  amount  if  the  interest  were  compounded  con- 

tinuously.    For  convenience  let  ,^,.    be  represented  by  —  •     Then 


the  amount,  y,  is1 

lira          TI        !  1^ 

y  =  a    1  +  -     100 

n  =  oo       L          u  J  , 

which  may  be  written 


u  - 

since  a,  r,  and  x  are  independent  of  n.    By  the  binomial  theorem, 
n        u(u  —  l)fln2 
wJ 


:m! 
LwJ 


lim 
n  =  oo 


.       C7  means  the  limit  of  U  as  n  increases  indefinitely. 


§103]          PERMUTATIONS,  COMBINATIONS,  ETC.  199 

Since  u  becomes  oo  as  n  becomes  oo ,  equation  (2)  gives 

i™  r1+i>=1  +  1+  i  +  i  +  i_  +  i  +  -     -(3) 

u  =  oo  L       uj  A       A  ° 

The  sum  of  the  infinite  series  on  the  right-hand  side  of  (3)  is 
represented  by  e  and  is  approximately  equal  to  2.71828.  It  is 
the  base  of  the  natural  or  hyperbolic  system  of  logarithms  (see 
footnote,  page  66).  Equation  (1)  then  reduces  to 

TX 

y  =  ae  Too.  (4) 

Since  r  is  any  constant,  r/100  is  any  constant,  call  it  6.  Then 
equation  (4)  may  be  written 

y  =  ae>*,  (5) 

where  a  and  b  are  constants.  It  is  the  equation  then  which 
expresses  the  law:  the  rate  of  growth  of  a  variable,  y,  is  directly  pro- 
portional to  the  variable  itself.  The  multiplier  a  is  the  value  of  the 
variable  when  time,  x,  is  zero,  and  6  is  the  proportionality  factor, 
i.e.,  it  expresses  the  ratio  of  the  rate  of  growth  of  y  to  y. 

The  compound  interest  law  is  one  of  the  very  important  laws  of 
nature. 

ILLUSTRATION:  Suppose  it  is  found  by  experiment  that  10  per- 
cent of  light  is  absorbed  in  passing  through  a  pane  of  glass  1  cm. 
thick.  Then  in  passing  through  a  second  pane  1  cm.  thick,  10 
percent  of  the  remaining  90  percent  will  be  lost,  and  so  on.  The 
intensity  of  light  in  passing  into  the  glass  obeys  then  the  com- 
pound interest  law, 

y  =  aebx 

where  y  represents  intensity  of  light  and  x  distance  into  the  glass. 
When  x  =  0,  y  =  a;  thus,  a  is  the  intensity  of  the  light  as  it  falls 
upon  the  glass  surface.  When  x  =  1,  y  =  0.9a,  hence 

0.9a   =  aeb, 
or 

0.9  =  eb 
or 

b  =  log«0.9 

Then  with  the  constants  a  and  6  determined,  the  intensity  of  the 
light  at  any  depth  below  the  surface  is  given  by 

aelog(0-9^  =  a(0.9)x 


CHAPTER  VIII 
PROGRESSIONS 

104.  Arithmetical  Progression  Defined.     An  arithmetical  pro- 
gression is  a  sequence  of  numbers  such  that  any  number  of  the 
sequence  except  the  first  may  be  obtained  from  the  preceding 
by  the  addition  (or  subtraction)  of  a  common  number,  called 
the     common    difference.     The     following     are     arithmetical 
progressions: 

1.  2,  4,  6,  8,  10,  12.  The  common  difference  is  2. 

2.  2,  5,  8,  11,  14.  The  common  difference  is  3. 

3.  —  7,  —  2,  3,  8,  13.  The  common  difference  is  5. 

4.  —  4,  —2,  0,  2,  4,  6.  The  common  difference  is  2. 

5.  2^,  2,  If,  1,  f,  0,  —  f.  The  common  difference  is  —1/2. 

Exercises 

From  the  following  sequences  select  those  which  are  arithmetical 
progressions,  and  give  the  common  difference: 

1.  5,  7,  9,  11,  13.  5.  x  -  y,  x,  x  +  y. 

2.  2,  4,  8,  16.  6.  x,  xy,  x  +  y. 

3.  10,  8,  6,  4,  2.  7.  0,  -  3,  -  6,  -  9. 

4.  3,  2f,  2i  2,  If.  8.  a,  a  +  d,  a  +  2d. 

105.  The  General  Term;  the  Sum  of  an  Arithmetical  Pro- 
gression.   Let  a  represent  the  first  term,  I  the  last  term,  d  the 
common  difference,  n  the  number  of  terms,  and  s  the  sum  of  all 
the  terms. 

The  general  series  is  then 

a,  a  +  d,  a  +  2d,  a  +  3d,  .    .    . ,  a  +  (w  —  l)d. 
It  is  obvious  that  the  coefficient  of  d  in  any  term  is  one  less  than 
the  number  of  the  term.    Thus,  in  the  fourth  term  it  is  3,  in  the 
twelfth  term  it  is  11,  and  in  the  nth  term  it  isn  —  1. 
A  formula  for  the  last  term  is  then 

1  =  a  +  (n  -  l)d.  (1) 

200 


§105] 


PROGRESSIONS 


201 


The  sum,  s,  is  found  by  adding  the  n  terms  together,  as 

s  =  a  +  (a  +  d)  +  (a  +  2d)  +  •        •  +  [a  +  (n  - 
or 

s  =  na+d[l  +  2  +  3  +  4+  •    .    .  +(ro-l)].  (2) 

The  sum  of  the  n  —  1  terms  written  within  the  brackets  may  be 
represented  graphically  as  follows:  Let  AoAn-i,  Fig.  94,  be  a  line 
n —1  units  long,  divided  into  n  —  1  equal  parts  by  the  points 
A i,  A2,  As,  etc.  Upon  the  segments  AoAi,  A\Az,  A2AZ,  .  .  ., 
An-zAn-i,  construct  rectangles  with  altitudes  respectively  1,  2,  3, 


Bo 


An.2  An.\ 


FIG.  94. 


4,  .  .  . ,  n  —  1  units  long.  The  areas  of  these  rectangles  are, 
respectively,  1,  2,  3,  4,  .  .  . ,  n  —  1  square  units.  The  areas  of 
these  rectangles  may  then  be  taken  as  representing  the  terms 
within  the  brackets  of  equation  (2) ,  and  the  sum  of  the  areas  of 
the  rectangles  representing  the  sum  of  the  terms  within  the 
brackets. 
Draw  A  0C. 

An-lC    =    AfiAn-1 

The  angle  CA0An.l  =  45° 

The  area  of  the  triangle  A^An^C  =  %(n  —  I)2 

The  area  of  each  shaded  triangle  above  the  line  A<>C  is  1/2. 


202  MATHEMATICS  [§105 

The  sum  of  the  n  —  1  triangles  is,  then,  %(n  —  1).     The  sum 
of  the  n  —  1  rectangles  is  then 

Kw-  1)2+  K»-  1)  =  *(«-  l)n. 
Therefore 

« 


Substituting  in  (2)  , 


TO  (TO  —  1) 
s  =  na-\  ----  -  d, 


or 

s  =  °[2a+(n-l)d]-  (3) 

From  formula  (3)  the  sum  of  an  arithmetical  progression  may 
be  calculated  if  the  first  term,  the  common-  difference,  and  the 
number  of  terms  are  given. 

Formula  (3)  may  be  written. 

*  *  2    a+  o+  (n  —  l)d    > 

or,  since  a  +  (n  —  l)d  =  I, 

'  s=  *(a+Z).  (4) 

ILLUSTRATION  1:  Find  the  sum  of  11,  22,  33,  .  .  .  to  twenty 
terms. 

This  is  an  arithmetical  progression,  for  any  term  is  eleven 
greater  than  the  next  preceding  term,  o  =  11,  d  =  11,  TO  =  20. 
Substituting  in  formula  (3), 

s  =  jpJ2fl  +  (n  -  l)d\  ='~  [2-11  +  (20  -  l)ll]  =  2310. 

ILLUSTRATION  2  :  Find  the  sum  of  \/2  +  2  \/2  +  3  \/2  4-  .  .  . 
to  eleven  terms. 

a  =  V%  d  =  \/2,  and  n  =  11. 

s  =  y[2A/2+  (11  -  1)V2J  =66  V2. 

ILLUSTRATION  3:  Find  the  sum  of  6  +  5i  +  4f  +  4  +  .  .  . 
to  fifty  terms. 

a  =  6,  d  =  —  f,  and  n  =  50. 


§107]  PROGRESSIONS  203 

Exercises 

1.  Find  the  sum  of  —  1  —  3£  —  6   .    .    .to  twelve  terms. 

2.  Find  the  sum  of  -£•  +  -5-  +  \  +   .    .    .to  eighteen  terms. 

3.  Find  the  sum  of  the  first  100  odd  numbers. 

4.  Find  the  sum  of  the  first  100  even  numbers. 

5.  A  man  saves  $200  each  year,  and  at  the  end  of  each  year  places 
this  amount  on  simple  interest  at  6  percent.     What  is  the  amount  of 
his  savings  at  the  end  of  ten  years? 

6.  For  drilling  a  well  a  contractor  is  to  receive  10  cents  for  the  first 
foot,  and  for  each  foot  thereafter  1/2  cent  more  than  for  the  pre- 
ceding foot.     What  does  he  receive  if  the  well  is  600  feet  deep? 

106.  Geometrical  Progression  Defined.     A   geometrical  pro- 
gression is  a  sequence  of  numbers  such  that  any  number  of  the 
sequence,  except  the  first,  may  be  obtained  from  the  preceding 
term  by  multiplying  by  a  common  number,  called  the  ratio. 
The  following  are  geometrical  progressions: 

1.  2,  4,  8,  16,  ...  The  ratio  is  2. 

2.  2,  -  4,  8,  -  16,  .    .    .  The  ratio  is  -  2. 

3.  2,  2^2,  4,  4  V2~  .    .    .  The  ratio  is  \/2. 

4.  "s/3,  -  3,  3  -v/3",  -  9,  ...  The  ratio  is  -  \/3, 

5.  a,  ar,  ar2,  ar3,  .    .    .  The  ratio  is  r. 

107.  The  Last  Term;  the  Sum  of  a  Geometrical Progresssion. 

Let  a  represent  the  first  term,  I  the  last  term,  r  the  ratio,  n  the 
number  of  terms,  and  s  the  sum  of  the  terms  of  a  geometrical 
progression. 

The  general  sequence  is,  then,  a,  ar,  ar2,  ar3,  .  .  . ,  ar*"1.  It 
is  obvious  that  the  exponent  of  r  in  any  term  is  one  less  than 
the  number  of  the  term.  Thus,  in  the  fourth  term  it  is  3,  in  the 
twelfth  term  it  is  11,  and  in  the  nth  term  it  is  n  —  1. 

A  formula  for  the  last  term  's  then 

1  =  ar"-1.  (1) 

The  sum,  s,  is  found  by  adding  the  n  terms  together,  as, 

s  =  a  +  ar  +  ar2  +  ar3  +  •    •    •  +  arn~l, 
or 

*  =  o(l  +  r  +  r2  +  r3  +  •  '•    •  +  r"-1).  (2) 


204  MATHEMATICS  (§107 

Since 

1  +  r  +  r2  +  r3  +  '    '    '  +  rn~l  =  1  ^r    » 

and  equation  (1)  becomes 

3.(.l      rnj  ff.\ 

s  =     t  _- —  (3) 

ILLUSTRATIONS:     1.  Sum  the  progression  2  +  4  +  8  +  •    •    • 
to  ten  terms. 

a  =  2,  r  =  2,  and  n  =  10. 

«d  ~  r-)  _  2(1  -  2iQ)  _ 
1-r  1-2 

2.  Sum  the  series  \/2  +  2  +  2  \/2  +  •    •   •  to  ten  terms. 

a  =  -\/2,  r  =  A/2,  and  n  =  10. 
V2[l  -  ( V2)10]  =  3lV2 
1  _  ^/2  V2  - 1 

3.  Sum  the  series  \/3  -  3  +  3  V3  -   •    •    •  to  ten  terms. 

a  =  A/3~,  ^  =  —  V3,  and  n  =  10. 


Exercises 

1.  Sum  3  +  9  +  27  +   •    •    •  to  ten  terms. 

2.  Sum  —   3+6—9+   •    •   -to  ten  terms. 

3.  Sum  3  —  1  +  1/3  —    •    •    •  to  six  terms. 

4.  Sum  —  3  +  6  +  15  +    •    •    -to  seven  terms. 

6.  Sum  \/3  +  2A/3  +  4  V3  +   •    •    -to  seven  terms. 

6.  Sum  \/3  +  2\/3~+  3\/3  +    •   •   •  to  seven  terms. 

7.  Sum  2  —  V2  +  1  —    •    •    •  to  six  terms. 

8.  Find  the  eleventh  term  in  each  of  the  above  series. 

1  An  exception  is  made  when  r  =  1,  but  in  this  case  1  +  r  +  r2  +  r3  + 
r""1  =  n. 


§1()S]  PROGRESSIONS  205 

9.  Show  that  the  only  right  triangles  whose  sides  are  in  arithmet- 
ical progression  are  those  whose  sides  are  proportional  to  3,  4,  and  5. 
HINT:    Let  x  —  k,  x,  and  x  +  k  represent  the  sides.     Find  k  in  terms 
of  x,  and  build  up  the  progression. 

10.  A  debt  of  $10,000  is  to  be  paid  in  ten  years.     An  equal  amount 
is  to  be  paid  at  the  end  of  each  year.     Find  this  amount  if  the  indebted- 
ness draws  interest  at  6  percent. 

11.  On  a  certain  twenty-year  life  insurance  policy  $32  premium 
is  paid  annually.     What  do  the  premiums  amount  to  by  the  end  of 
the  twentieth  year,  interest  compounded  annually  at  3^  percent? 

12.  An  equal  amount  of  money  is  deposited  at  the  end  of  each 
year  for  twenty  years  as  a  sinking  fund  to  replace  a  piece  of  machinery 
valued  at  $10,000.     How  much  must  be  deposited  each  year  if  the 
deposits  draw  4  percent  compound  interest? 

108.  The  Infinite  Geometrical  Progression.  From  the  formula 
for  the  sum  of  a  geometrical  progression,  s  =  — ^ ,  it  is 

evident  that,  if  r  is  greater  than  unity  in  numerical  value,  the 
numerical  value  of  s  increases  without  limit  as  r  increases  without 
limit.  For,  as  n  grows  very  large,  rn  grows  large  without  limit, 
causing  s  to  grow  large  without  limit. 

If  r,  however,  is  numerically  less  than  unity,  rn  grows  smaller 
and  smaller  as  n  increases,  and  approaches  zero  as  n  increases 
without  limit.  Thus,  the  quantity  within  the  parentheses 
approaches  1,  or  the  limit  of  s  is  a/(l  —  r)  as  n  increases  without 
limit.  The  sum,  s,  of  a  geometrical  progression  approaches  a 
definite  finite  limit  as  n  increases  without  limit,  when  r  is  nu- 
merically less  than  unity.  This  limit,  a  /(I  —  r),  is  called  the 
sum  of  the  infinite  geometrical  progression. 

ILLUSTRATIONS:  1.  Find  the  sum  of  the  infinite  progression 
1  +  £  +  i  +  i  +  '  •  •  to  oo. 

a  =  1  and  r  =  %. 

*          -o 

•  1-r      1-t 

2.  Find  the  sum  of  the  infinite  progression  \/3  +  1+1  \/3  — 
'  to  0°. 

a  =  V3  and  r  =  I/  \/3. 

a  \/3  3         =  3(V'3  +  1)< 

\/3  -  1  =  2 


206  MATHEMATICS  [§108 

3.  Express  the  repeating   decimal  0.237237237  ...  in  frac- 
tional form. 

237  237 

The  decimal  fraction  0.237237237  .  .  .  means  ^^  +  (1QOQ)  2 

237 

+  •    •    •,  an  infinite  geometrical  progression,  in  which 


a  =  237  /1000,  and  r  =  1  /1000.     Substituting  in  the  formula  for 
the  sum  of  an  infinite  geometrical  progression, 

237 

1000      237     237 
1    1000  -  1  ==  999' 
~  1000 

Exercises 

1.  Find  the  sum  ofl—  £  +  i  —  |+    .    .    .  to  <» . 

2.  Find  the  sum  of  \/2  —  ?\/6  +  i\/2  —    .    .    .  to  °° . 

3.  Find   the    sum    of — 7= -^  +  1  —  — 7=-     — ;=  + .        .  to  oo  . 

\/2  +  \/3  \/2  -  \/3 

4.  Express  0.7261872618   ...  in  fractional  form. 

6.  The  middle  points  of  the  sides  of  a  square,  10  inches  on  a  side, 
are  joined  forming  a  smaller  square.  The  middle  points  of  the  sides 
of  this  square  are  joined  forming  a  third  square.  This  process  of  form- 
ing smaller  squares  is  continued  indefinitely.  Find  the  sum  of  the 
perimeters  of  the  squares. 

6.  Same  as  exercise  5,  but  use  an  equilateral  triangle  10  inches  on 
a  side. 


CHAPTER  JX 
'[PROBABILITY] 

109.  Probability  Defined.  When  a  coin  is  thrown  upon  the  flat 
surface  of  a  table,  it  falls  in  one  of  two  ways,  heads  or  tails.  If  it 
is  given  a  whirling  motion  when  thrown,  the  chances  of  its  falling 
heads  are  no  greater  than  the  chances  of  its  falling  tails,  and  the 
chances  of  its  falling  tails  are  no  greater  than  the  chances  of  its 
falling  heads;  that  is,  before  the  coin  fell,  we  had  no  reason  for 
expecting  one  rather  than  the  other.  All  that  could  be  said  was 
that  the  coin  must  fall  in  one  of  two  ways,  both  equally  likely. 

When  a  common  die  is  thrown  it  falls  in  one  of  six  possible  ways, 
all  equally  likely.  One  and  only  one  of  the  six  ways  is  that  of  the 
ace  up.  The  chances  then  of  throwing  an  ace  with  one  throw  of  a 
die  is  one  in  favor  and  five  against,  or  one  to  five,  or  one  out  of 
six.  This  same  thought  is  expressed  by  saying  that  the  probability 
of  throwing  an  ace  with  one  throw  of  a  die  is  one  to  six,  or  1  /6. 

If  a  thing  can  happen  in  n  ways,  and  can  fail  to  happen  in  m 
ways,  it  can  happen  or  fail  in  just  m  -\-  n  ways.  We  say  there 
are  m  +  n  possible  cases,  m  favorable  cases,  and  n  unfavorable 

cases.  If  all  these  ways  are  equally  likely  to  occur,  — ^p —  (the 
ratio  of  the  favorable  cases  to  the  possible  cases)  is  called  the 
probability  of  the  event  happening.  — j- —  (the  ratio  of  the 

unfavorable  cases  to  the  possible  cases)  is  called  the  probability 
of  it  failing. 

When  a  coin  is  thrown,  the  probability  of  heads  turning  is 
1  /2;  for  the  number  of  favorable  cases  is  one,  and  the  number  of 
possible  cases  is  two.  The  probability  of  heads  not  turning  is  1  /2. 

When  a  die  is  thrown,  the  number  of  possible  cases  is  six,  the 
number  of  favorable  cases  of  the  ace  falling  up  is  one.  The  proba- 
bility of  throwing  an  ace  is  then  1/6.  The  probability  of  not 
throwing  an  ace  is  5  /6. 

207 


208  MATHEMATICS  [§109 

In  a  set  of  twenty-eight  dominoes  there  are  seven  doublets. 
When  one  domino  is  selected  at  random  the  probability  of  its 
being  a  doublet  is  7/28,  or  1/4.  For,  the  selection  of  one 
domino  may  be  made  is  twenty-eight  ways,  or  the  number  of 
possible  cases  is  twenty-eight;  and  since  a  doublet  may  be 
selected  in  seven  ways,  the  number  of  favorable  cases  is  seven. 

When  two  dice  are  thrown  the  number  of  ways  in  which  they 
may  fall  is  6  X  6  =  36.  Of  these  thirty-six  ways,  one,  and  one 
only,  is  a  pair  of  aces.  The  probability  of  throwing  a  pair  of  aces 
with  two  dice  is  then  1  /36.  The  probability  of  throwing  a  pair 
is  6/36,  or  1/6.  The  probability  of  throwing  a  sum  less  than 
5  is  1  /6. 

If  an  event  is  certain  to  happen,  the  number  of  favorable  cases 
is  equal  to  the  number  of  possible  cases,  and  the  probability  of 
its  happening  is  one.  If  an  event  is  certain  to  fail  the  number  of 
favorable  cases  is  zero,  and  the  probability  of  its  happening  is 
zero.  In  all  other  cases  the  probability  of  an  event  happening 
is  a  positive  proper  fraction. 

If  an  event  may  happen  in  a  ways,  and  fail  in  b  ways,  the 

Q 

probability  of  its  happening  is  -  — r»  and  the  probability  of  its 

a.  ~T~  D 

failing  is  — 7— r-«       The  sum  of  these  two  probabilities  is  1. 


Exercises 

1.  Two  dice  are  thrown.     What  is  the  probability  of  throwing  a 
5  and  a  3?     A  sum  greater  than  7?     A  sum  equal  to  10?     Conse- 
cutive numbers?     One  and  only  one  6? 

2.  A  box  contains  three  white,  two  black,  and  five  red  balls;  if  one 
is  drawn,  what  is  the  probability  it  is  white?     Black?    Red?     If  two 
are  drawn,  what  is  the  probability  that  both  are  white?     Both  black? 
Both  red?     One  white  and  one  black?     One  white  and  one  red? 
One  black  and  one  red? 

3.  Three  coins  are  thrown.     What  is  the  probability  that  all  turn 
heads?     That  two  turn  heads?     That  only  one  turns  heads? 

4.  In  a  box  there  are  five  red,  two  white,  and  seven  blue  balls. 
If  three  are  drawn  one  after  the  other,  what  is  the  probability  of 
their  being  drawn  in  the  order  red,  white,  and  blue? 


§110]  PROBABILITY  209 

HINT:  The  total  number  of  ways,  in  order,  in  which  three  balls 
can  be  drawn  from  fourteen  balls  is  14P  3  . 

6.  Four  cards  are  drawn  from  a  pack.  What  is  the  probability 
that  there  is  one  from  each  suit? 

110.  Mutually  Exclusive  Events.  Two  or  more  events  are  said 
to  be  mutually  exclusive  if  the  occurrence  of  one  excludes  the 
occurrence  of  the  others.  Thus,  a  5  and  a  6  cannot  be  thrown 
together  with  one  die.  The  occurrence  of  one  event  excludes  the 
occurrence  of  the  other.  The  two  events  are  mutually  exclusive. 

Let  A  and  B  be  two  mutually  exclusive  events.  There  are  three 
possible  cases: 

(1)  The  event  A  may  happen  and  the  event  B  fail. 

(2)  The  event  B  may  happen  and  the  event  A  fail. 

(3)  The  event  A  and  the  event  B  may  both  faiJ. 

If  a,  b,  and  c  are,  respectively,  the  number  of  equally  likely 
possible  cases,  the  probability  that  either  A  or  B  happens  is 

— .     For  a  +  6  +  c  is  the  number  of  possible  cases,  and 
a  +  o  +  c 

a  +  b  is  the  number  of  favorable  cases.     The  probability  that  A 

happens  is  —  — ,  and  that  B  happens  is  —  —  .      Hence 

a  +  6  +  c'  a  +  6+c 

the  probability  of  the  occurrence  of  one  or  the  other  of  two  mutually 
exclusive  events  is  the  sum  of  the  probabilities  of  the  two  separate 
events. 

It  is  easily  seen  that  this  law  may  be  extended  to  any  number 
of  mutually  exclusive  events.  The  above  law  applies  only  to 
mutually  exclusive  events,  and  care  must  be  exercised  not  to 
apply  it  to  events  not  so  related.  Thus,  if  A's  probability  of 
winning  a  contest  is  1  /3,  and  if  B's  probability  of  winning  in  the 

same  contest  is  1  /4,  the  probability  of  A  or  B  winning  is  5-  +  j-  = 

o       4 
7 
^.    For  the  two  events  are  mutually  exclusive.     If  A's  probability 

LZi 

of  solving  a  problem  is  1  /3,  and  B's  probability  of  solving  the  same 
problem  is  1  /4,  the  probability  of  the  problem  being  solved  is  not 
the  sum  of  the  separate  probabilities,  for  the  events  are  not 
mutually  exclusive.  The  solution  of  the  problem  by  A  does  not 
prevent  B  from  solving  it. 

14 


210  MATHEMATICS  [§111 

Exercises 

1.  The  probability  of  throwing  an  ace  with  a  common  die  is  |; 
the  probability  of  throwing  a  6  is  f .     What  is  the  probability  of 
throwing  an  ace  or  a  6? 

2.  A  ticket  is  drawn  from  a  set  numbered  from  1  to  50.    What 
is  the  probability  that  it  is  a  multiple  of  either  7  or  9? 

3.  A  die  is  thrown.     What  is  the  probability  that  the  throw  will 
be  greater  than  3? 

4.  Two  dice  are  thrown.     What  is  the  probability  that  the  throw 
will  be  greater  then  7? 

5.  A  bag  contains  four  red,  five  white,  two  blue,  and  seven  green 
balls.     One  is  drawn.     What  is  the  probability  that  it  is  either  red  or 
blue? 

111.  Independent  Events.  The  occurrence  of  several  single 
events  simultaneously  or  in  succession  constitutes  a  compound 
event.  Events  are  said  to  be  dependent  or  independent,  if  the 
occurrence  of  one  does  or  does  not  affect  the  occurrence  of  the 
others. 

Thus,  the  drawing  of  a  red  and  a  white  ball  from  a  bag  containing 
red,  white,  and  blue  balls  is  a  compound  event. 

Of  two  independent  events,  let  a  be  the  number  of  ways,  all 
equally  likely,  the  first  may  happen,  and  let  b  be  the  number  of 
ways,  all  equally  likely,  it  may  fail;  let  ai  be  the  number  of  ways, 
all  equally  likely,  the  second  may  happen,  and  let  61  be  the  number 
of  ways,  all  equally  likely,  it  may  fail.  The  number  of  possible 
cases  for  the  first  event  is  a  +  6,  and  the  number  of  possible  cases 
for  the  second  event  is  a\  -\-  b\.  Each  case  of  the  first  event  may 
be  associated  with  each  case  of  the  second  event.  The  number  of 
possible  cases  of  the  compound  event  is  then  (a  +  b)(a\  +  &0, 
all  equally  likely  to  occur.  In  aa\  cases  both  events  happen; 
in  661  cases  both  events  fail;  in  061  cases  the  first  event  happens  and 
the  second  event  fails;  and  in  a\b  cases  the  first  event  fails  and  the 
second  event  happens.  Then  the  probability 

of  both  events  happening  is  -, — 


of   both    events    failing    is  -. — ,   ,.,  1    ,   ,  .  > 

(a  +  6)(ai  +  61) 


§111]  PROBABILITY  211 

of  the  first  happening  and  _  061  _  t 
second  failing  is  (a  +  &)(fli  +  &i) 

of  the  first  failing  and  second  _  Oi&  _  _ 
happening  is  (a 


From  the  first  probability  given  above  it  is  seen  that  the 
probability  of  the  occurrence  of  two  independent  events  is  the  product 
of  the  probabilities  of  the  occurrence  of  the  separate  events. 

This  law  may  be  extended  to  include  three  or  more  single 
events. 

ILLUSTRATIONS:  1.  A  bag  contains  three  white  and  four  black 
balls.  The  probability  of  drawing  a  white  ball  is  3  /7.  If  a 
white  ball  is  drawn  and  not  replaced,  the  probability  of  drawing 
another  white  ball  is  2/6.  The  probability  of  drawing  two  white 

u  11    •  •      •    3  ^  1       l 

balls  in  succession  is  ^  X  ^  =  ^' 
lot 

2.  If  A's  probability  of  solving  a  problem  is  1  /3,  and  if  B's  proba- 
bility of  solving  it  is  l'/4,  the  probability  that  A  does  not  solve  it 
is  2  /3,  and  the  probability  that  B  does  not  solve  it  is  3  /4.     The 

231 

probability  that  both  fail  to  solve  the  problem  is  ^  X  7-  =  «"• 

Since  1  /2  is  the  probability  the  problem  is  not  solved,  1  —  1/2,  or 
1  /2,  is  the  probability  that  the  problem  will  be  solved  by  A  or  B. 

3.  Five  coins  are  thrown.     What  is  the  probability  that  they 
will  all  turn  heads?     The  probability  that  a  coin  turns  heads  is  1  /2. 

Hence  the  probability  that  the  five  turn  heads  is  o'o'o'o'o  =Q2' 

Exercises 

1.  What  is  the  probability  that  at  least  one  ace  will  turn  when 
three  dice  are  thrown? 

2.  A  bag  contains  six  white  and  five  red  balls,  and  a  second  bag 
contains  four  white  and  three  black  balls.     A  ball  is  drawn  from  each, 
what  is  the  probability  that  both  are  white?     That  one  is  white  and 
one  black?     That  one  is  white  and  one  red? 

3.  What  is  the  probability  of  holding  four  aces  in  a  game  of  whist? 

4.  A  sack  contains  three  white  and  two  black  balls;  a  second  sack 
contains  four  white  and  seven  black  balls.     A  ball  is  drawn  from  the 
first  sack  and  placed  in  the  second;  then  a  ball  is  drawn  from  the 


212  MATHEMATICS  [§112 

second  sack.     What  is  the  probability  that  the  second  ball  drawn  is 
white? 

5.  A  bag  contains  three  white  and  four  black  balls.  Two  persons 
draw  alternately  one  ball  without  replacing  it.  If  A  draws  first, 
what  is  the  probability  that  he  is  the  first  to  draw  a  white  ball? 

112.  Expectation.     If  p  denotes  the  probability  of  a   person 
winning  a  sum  of  money,  m,  the  product  mp  is  called  the  value 
of   his    expectation,   or   his   mathematical    expectation.    Thus, 
if  a  man  is  to  receive  $10  if  he  draw  a  doublet  from  a  set  of 
twenty-eight  dominoes  with  one  trial,  his  mathematical  expec- 
tation  is  $2.50,  found  by  multiplying  $10  by  1/4,  the  proba- 
bility of  drawing  a  doublet. 

113.  Successive  Trials.     On  a  single  trial,  let  p  be  the  proba- 
bility of  an  event  happening  and  let  q  be  the  probability  of  the 
event  not  happening.     Thus,  if  a  die  is  thrown  but  once,  the 
probability  of  throwing  an  ace  is  1/6  =  p,  and  the  probability 
of  not  throwing  an  ace  is  5/6  =  q.     If  the  die  is  thrown  twice, 
the  probability  of  throwing  an  ace  each  time  is  p2,  or  (1  /6)2  =  1  /36. 
The  probability  of  not  throwing  an  ace  is  q2  —  25  /36.     The 
probability  of  throwing  an  ace  on  the  first  trial  and  not  throwing 

155 

an  ace  on  the  second  trial  is  pq,  or  «  X  ^  =  ™"     The  probability 

o      o      oo 

of  not  throwing  an  ace  on  the  first  trial  and  throwing  an  ace  on 

515 

the  second  trial  is  qp,  or  _•  X  ^  =  ~a'    The  probability  of  throw- 
ing an  ace  once  and  once  only  with  two  trials  is  ™  +  ™  =  j^- 

oo       oo       oo 

The  probability  of  throwing  an  ace  at  least  once  with  two  trials 

'     l°--l-  —  =  — 

18  36  +  36  ~  36* 

The  probability  of  the  event  happening  on  each  of  the  two 
trials  is  p2.  The  probability  of  the  event  happening  on  the  first 
trial  and  not  happening  on  the  second  trial  is  pq.  The  proba- 
bility of  the  event  not  happening  on  the  first  trial  and  happening 
on  the  second  trial  is  qp.  The  probability  of  the  event  happen- 
ing but  once  is  2pq.  The  probability  of  the  event  not  happening 
upon  either  of  the  trials  is  q2. 

p*  +  2pq  +  <?  =  1 


§113]  PROBABILITY  213 

In  the  illustration  above, 

P'  +  2W  +  ««-Sg  +  SJ  +  |-!-» 

The  probability  of  the  event  happening  on  each  of  three  trials 
is  p3.  The  probability  of  it  happening  on  the  first  two  trials 
and  not  happening  on  the  last  trial  is  p'2q;  of  happening  on  the 
first  and  last  trials  and  not  happening  on  the  second  trial  is 
p2q;  and  of  happening  on  the  last  two  trials  and  not  happening 
on  the  first  trial  is  pzq.  The  probability  of  the  event  happening 
twice  and  twice  only  with  three  trials  is  3p2g.  The  probability 
of  the  event  happening  once  and  once  only  with  three  trials  is 
3pg2.  The  probability  of  the  event  failing  to  happen  with  three 
trials  is  q3.  The  probability  that  the  event  happens  at  least 
once  with  three  trials  is  p3  +  3pzq  +  3pqz. 

p3  +  3p2g  +  3pq2  +  q3  =  1 

A  sack  contains  three  white  balls  and  seven  black  balls. 
Three  successive  draws  of  one  ball  are  made,  each  ball  being 
returned  to  the  sack  before  the  next  draw  is  made,  p  =  3  /10 
and  q  =  7  /10.  The  probability  of  drawing  a  white  ball  each 
time  is  (3/lQ)3  =  27/1000;  of  drawing  two  white  balls  and  one 
black  ball  is  3(3/10)2(7/10)  =  189/1000;  of  drawing  one  white 
ball  and  two  black  balls  is  3(3/10) (7 /10)2  =  441/1000;  and  of 
drawing  a  black  ball  each  time  is  (7/10)3  =  343/1000.  The 
probability  of  drawing  a  white  ball  at  least  once  is 

27         189        441         657 

I       -|  f\f\f\      I 


1000  '  1000  '  1000   1000 

If  n  trials  are  made  the  probability  of  the  event  happening 
on  each  trial  is  pn.  The  probability  of  the  event  happening  n  —  1 
times  and  failing  to  happen  one  time  is  npn-1q.  For,  the 
probability  of  its  failing  to  happen  on  the  last  trial  only  is  pn~lq, 
of  its  failing  to  happen  on  the  next  to  the  last  trial  only  is  pn~lq, 
of  its  failing  to  happen  on  the  second  from  the  last  trial  only 
is  pn~lq,  and  so  on  for  the  n  trials  on  which  the  event  may  fail 
to  happen;  then  the  probability  of  the  event  failing  to  hap- 
pen one  time  only  in  the  n  trials  is  the  sum  of  these  probabilities, 
or  np"-^.  In  a  similar  way  it  is  shown  that  the  probability 


214 


MATHEMATICS 


[§113 


of  the  event  happening  on  n  —  2  trials  and  failing  to  happen 
on  two  trials  is  nC2pn~zqz  (where  nC2  is  the  number  of  combina- 
tions of  n  things  taken  two  at  a  time),  or 


n  —  2 


The  probability  that  the  event  happens  on  n  —  3  trials  and  fails 
to  happen  on  three  trials  is = pn-3qs.     The  probability 


3    n-3 


that  the  event  happens  on  n  —  r  (r 
n 


on  r  trials  is 


n)  trials  and  fails  to  happen 


n  —  r 


The  probability  that  the  event 


happens  on  each  trial,  on  each  trial  but  one,  on  each  trial  but  two, 
on  each  trial  but  three,  etc.,  are,  respectively,  the  first,  second, 
third,  etc.,  terms  of  the  binomial  expansion, 


(p  +  q)n  =  P"  +  np-'-'q  + 


=_  pn-2q2 


2     n-2 


3  I  n-3 


n  —  r 


•  +qn  = 


The  probability  that  the  event  happens  on  k  trials  and  on  k 
trials  only  is = p*qn~fc.     The  probability  that  the  event 


happens  on  at  least  k  trials  out  of  n  trials  is 


npn-1q 


2     n  -  2 


k-1    k+1 


n  -k     k 


Thus,  if  a  die  is  thrown  six  times,  the  probability  of  throwing 
an  ace  three  times  and  three  times  only  is 
6 


3      6-3 


186,624 


Exercises 


1.  What  is  the  probability  of  throwing  exactly  three  6's  in  five 
throws  with  a  single  die? 

2.  The  odds  are  2: 1  in  favor  of  A  winning  a  single  game  against  B. 
Find  the  probability  of  A  winning  at  least  two  games  out  of  three. 


§114] 


PROBABILITY 


215 


[114.]  Graphic  Representation  of  the  Terms  of  (p  +  q)n.  Let 
A  and  B  represent,  respectively,  the  (r  —  l)st  and  rth  terms  of  the 
expansion  of  (p  +  q)n~l,  and  let  C  represent  the  rth  term  of  the 
expansion  of  (p  +  q)n.  Then: 

n-  1 

A  =  n_iCr_ipn~i+V~2  =  i —  r~r"  pn~r+1gr~2, 

r  —  2      n  —  r  +  1  ^ 


n-  1 


r-  1 


C  —         r     ,  T)n-r+l,,r—  1  — 

»c'-iP        9       - 


n  —  r 


pn-rqr-l} 


Y- 


From  these  expressions  it  will  be  shown  that 
C  =  Aq  +  Bp. 


Aq  = 


Bp  = 


n-l 


r-2      n-r+1 


n-l 


r  -  1 


n  —  r 


and 


-l    L        -L- 

—  L r ~ 2   n-r  +  1 


-J_    n  -r  |: 


or 


Ap  +  Bq 


=  I71"1    I  -r  ~  1 

I   T —  1     n  —  r+1 


+ 


or 


Ap  +  Bq  = 


r  -I  In-r+1 


'-1  =  C. 


216 


MATHEMATICS 


[§114 


The  general  term,  the  rth  term,  of  the  expansion  of  (p  +  q)n  may 
be  found  from  the  (r  —  list  and  rth  terms  of  the  expansion  of 

(p  +  g)"-1- 

This  property  of  the  binomial  expansion  makes  it  possible  to 
construct  graphically  the  terms  of  the  expansion  for  the  case 


FIG.  95. 

p  +  q  =  1.  Let  AA'  and  BB',  Fig.  95,  whose  lengths  represent, 
respectively,  A  and  B,  be  drawn  perpendicular  to  AB.  CC'  is  a 
third  perpendicular  drawn  such  that 

AC  =  GE_ 
p    '      q  ' 

Draw  AB'  cutting  CC'  at  K.  From  the  similar  triangles  AA'B' 
and  KC'B', 

KG'      AA' 


or 


CB  '    AB 

AA'  X  CB 


KG'  = 


AB 


(1) 


From  the  similar  triangles  BB' A  and  CKA, 
CK  _  BB' 

AC~  AB' 


§114]  PROBABILITY  217 

or 


BB'  X  AC 

AB      ' 


By  adding  (1)  and  (2), 


,  BB'XAC  +  AA'XCB 

AB  ~ 


or 


_  B(AC)  +  A(CB)  _  B(AC)  +  A(CB) 
AB  AC  +  CB      ' 

or 


cc, 


or,  since  p  +  q  =  1,  <7C'  =  JBp  +  Ag.  Therefore  <7C'  =  C, 
the  rth  term  of  the  expansion  (p  +  5)",  when  p  +  q  =  1. 

The  first  term  of  the  expansion  of  (p  +  q)n  equals  the  first 
term  of  the  expansion  of  (p  +  q)n~l  multiplied  by  p,  and  may  be 
found  graphically  as  in  Fig.  95  by  taking  AA'  zero.  The  last 
term  of  the  expansion  of  (p  +  q)n  equals  the  last  term  of  the 
expansion  of  (p  +  g)""1  multiplied  by  q,  and  may  be  found 
graphically  as  in  Fig.  95  by  taking  BB'  zero. 

Let,  0,  1,  2,  3,  4,  5,  6,  7,  .  .  .  ,  Fig.  96,  be  points  chosen 
equidistant  along  a  horizontal  line.  Let  C\,  C*,  Ca,  Ct,  .  .  . 
be  intermediate  points  chosen  such  that 

p       OCi      1C2      2C3 

q  ~  Cil  ~  C£  ~  C33  " 

At  these  points  of  division  of  the  line  07  erect  perpendiculars 
and  let  these  perpendiculars  be  designated  by  the  letter  or  numeral 
at  its  lower  end.  We  shall  show  how  to  lay  off,  by  means  of  a 
graphical  construction,  the  successive  terms  of  the  expansion  of 
(p  +  q)n.  With  some  convenient  scale  let  IP  =  p  and  2Q  =  q. 
Then  IP  and  2Q  represent  the  terms  of  the  expansion  when  n  =  1. 
Draw  OP,  PQ,  and  0.3.  OP  cuts  Ci  at  ai,  PQ  cuts  C2  at  0lf  and 
0,3  cuts  C3  at  71.  cti  projected  upon  1  gives  01,  /Si  projected  upon 


218 


MATHEMATICS 


[§114 


2  gives  61,  and  71  projected  upon  3  gives  Ci.  By  what  was  shown 
in  the  first  part  of  this  section  it  will  be  seen  that  la\,  2b\,  and  3ci 
represent  the  terms  of  (p  +  q)z.  In  a  similar  manner  it  is  shown 
that  laz,  262,  3c2,  and  4d2  represent  the  terms  of  the  expansion 
(p  +  <?)3;  1«3,  263,  3c3,  4d3,  and  5e3  represent  the  terms  of  (p  +  g)4; 
Ia4,  264,  3c4,  4d4,  504,  and  6/714  represent  the  terms  of  (p  +  <?)8. 
Continuing  in  this  way  the  terms  of  the  expansion  could  be  repre- 
sented for  any  positive  integral  value  of  n. 


c,    \ 

ft 


O        Ci  1        C2  2 


3        C4  4         Cs  5 

FIG.  96. 


C«  6 


C,  7 


If  p  +  g  is  not  1,  but  k,  the  expression  (p  +  q)»  could 
be  written  (kp'  +  /e?')re,  or  fcn(w'  +  q')n,  where  p'  +  q'  =  1. 
The  terms  of  (p'  +  g')n  may  now  be  represented  as  above  and 
finally  all  ordinateo  increased  in  the  ratio  of  1 :  fc",  or  the  vertical 
scale  changed  so  that  all  vertical  distances  represent  quantities 
kn  times  as  great  as  originally. 

If  p  —  q,  the  configuration,  Fig.  96,  will  be  symmetrical  with 
respect  to  some  vertical  line.  In  this  case  (p  =  q  =  1/2)  the 
ordinates  represent  the  binomial  coefficients.  When  p  ^  q 
the  configuration  is  said  to  be  asymmetrical  or  skew. 


§114] 


PROBABILITY 


219 


In  Fig.  97  the  ordinates  represent  the  binomial  coefficients  of 
the  999th  power  of  (p  +  q).    The  drawing  is  due  to  Quetelet.1 


FIG.  97. 

From  what  was  shown  in  the  first  part  of  this  section  the  binomial 
coefficients  of  any  power  may  be  calculated  from  those  of  the 
expansion  of  a  power  one  lower. 


n 

Binomial 

Coefficients 

I 

1 

1 

2 

1 

2 

1 

3 

1 

3 

3 

1 

4 

1 

4 

6 

4 

1 

5 

1 

5 

10 

10 

5 

1 

6 

1 

6 

15 

20 

15 

6 

1 

7 

1 

7 

21 

35 

35 

21 

7         1 

Thus,  in   the  above,    any  number  is  the  sum   of  the  number 
directly  above  it  and  the  number  to  the  left  of  the  latter. 


Exercises 

1.  Represent   graphically,   as  in  Fig.  96,  the  terms  of   (p  +  q)n, 
if  p  =  \,  q  =  -J,  and  n  =  7. 

1  The  limit  of  the  broken  line  at  the  top  of  the  ordinates  is  known  as  the  probability 
curve;  its  equation  is  of  the  form 

V  -  a«"kV 
as  is  shown  in  treatises  on  the  Theory  of  Probability. 


220 


MATHEMATICS 


[§115 


2.  Represent  graphically,  as  in  Fig.  96,  the  terms  of  (p  +  q)n, 
if  p  =  q  =  5  and  n  =  7. 

116.  Deviations  from  the  Arithmetic  Mean  and  the  Probability 
Curve.  The  readings  of  a  set  of  direct  measurements  of  a 
magnitude  (as  the  elevation  of  a  bench  mark,  the  length  of  a 
base  line,  the  angle  subtended  by  a  distant  object,  etc.)  taken 
by  the  same  person  or  by  several  persons,  will,  in  general,  disagree 
among  themselves.  This  is  especially  true  if  the  measuring 
instrument  is  such  as  to  enable  the  observer  to  read  to  a  small 
part  of  the  unit;  for  example,  as  the  reading  of  a  level  rod  to  the 
nearest  one-thousandth  of  a  foot. 


6.5  7.0  7.5  8.0  8.5 

FIG.  98. — A  frequency  distribution  polygon. 


9.0 


It  is  assumed  that  the  measuring  instruments  are  in  adjust- 
ment and  that  proper  care  is  exercised  in  handling  and  reading 
them.  The  disagreements  of  the  readings  are  then  due  to  what 
are  known  as  errors  of  observation. 

In  Table  XIV  are  given  readings  of  the  area  bounded  by  an 
irregular  curve  drawn  at  random.  The  readings  were  taken  with 
an  Amsler  Polar  Planimeter.  In  column  I  are  given  the  readings 
of  the  instrument  and  in  column  II  are  given  the  number  of  times 
the  corresponding  reading  was  obtained.  Fig.  98  gives  a  graphic 
representation  of  the  distribution  of  the  frequency  of  the  readings. 
Areas  are  represented  by  abscissas  and  frequencies  of  readings  by 
ordinates.  To  represent  more  clearly  to  the  eye  the  distribution 
of  the  frequencies  of  the  readings,  the  plotted  points  may  be 
joined,  in  order,  by  straight  lines;  or  through  each  point  a  hori- 
zontal line  may  be  drawn  giving  a  series  of  rectangles,  as  is 
illustrated  by  the  figure.  The  length  of  the  base  of  each  rectangle, 


§115] 


PROBABILITY 


221 


in  this  case,  represents  the  least  count  of  the  instrument,  while 
its  altitude  represents  the  frequency  of  a  reading. 


TABLE  XIV 


I. 

II. 

ill 

IV 

V 

VI 

Area, 
sq.  cm. 

No.  of  read- 
ings 

/ 

» 

A1 

126.4 

0 

1 

-12.6 

158.76 

126.5 

1 

4 

-11.6 

538.24 

126.6 
126.7 

3 
4 

8 
13 

-10.6 
-  9.6 

898.88 
1198.08 

8 

126.8 

6 

17 

-  8.6 

1257.32 

126.9 
127.0 

7 
9 

22 
31 

-  7.6 
-  6.6 

1270.72 
1350.36 

22 

127.1 

15 

42 

—  5  6 

1317  12 

127.2 
127.3 

18 
22 

55 
69 

-  4.6 
-  3.6 

1163.80 
894.24 

55 

127.4 

29 

84 

-  2.6 

567.84 

127.5 
127.6 

33 
31 

93 
96 

-   1.6 
-  0.6 

238.08 
35.56 

93 

127.7 

32 

92 

0.4 

14.72 

127.8 
127.9 

29 

28 

89 
81 

1.4 
2  4 

150.41 
465.56 

89 

128.0 

24 

67 

3.4 

774.52 

128.1 
128  2 

15 
12 

51 

37 

4.4 
5  4 

987.36 
1078  92 

51 

128.3 

10 

31 

6  4 

1269.76 

128.4 
128  5 

9 

7 

26 
21 

7.4 
8  4 

1423.76 
1481  76 

26 

128.6 

5 

13 

9.4 

1148.68 

128.7 
128.8 

1 
2 

8 
5 

10.4 
11  4 

865.28 
649  80 

8 

128.9 

2 

4 

12  4 

615.04 

129.0 

0 

2 

13.4 

359.12 

2 

Sums  .  . 

354 

1062 

22,173.68 

Mean  area     =  127.66  sq.  cm. 
h2   =  0.024 

4=  =  0.087 
Vtr 


<r  =  4.3 
E,.r  =  3.08  or  0.308  sq.  cm. 

E  =  0.1  or  0.01  sq.  cm. 


222  MATHEMATICS  [§115 

The  graph  obtained  by  either  method  of  connecting  the  plotted 
points  is  called  a  frequency  distribution  polygon. 

From  Fig.  98,  as  well  as  from  the  table  of  readings,  one  is  lead 
to  believe  that  the  true  area  sought  lies  somewhere  between 
127.3  and  128.0,  probably  near  127.6  square  centimeters. 

If  it  be  granted  that  in  the  long  run  numerically  small  errors 
occur  more  frequently  than  larger  ones,  and  that  positive  and 
negative  errors,  numerically  equal,  occur  with  equal  frequency, 
the  frequency  polygon  for  a  very  large  number  of  readings  would 
present  symmetry  with  respect  to  a  vertical  line,  and  the  altitudes 
of  the  rectangles  would  decrease  with  increasing  distances  from 
this  line  of  symmetry.  In  such  a  case  the  altitude  of  any  rec- 
tangle, or  its  area,  would  be  proportional  to  the  probability  that 
a  single  reading  fall  within  the  limits  defined  by  the  ends  of  its 
base.  For,  the  sum  of  the  altitudes  of  all  rectangles  would 
represent  the  total,  or  possible,  number  of  cases,  while  the  altitude 
of  a  particular  rectangle  would  represent  the  number  of  cases  of 
the  corresponding  reading.  Since  the  bases  of  the  rectangles  are 
equal,  the  above  statement  holds  for  the  areas  of  the  rectangles  as 
well  as  for  their  altitudes. 

If  the  scales  of  the  drawing  are  so  chosen  that  the  sum  of 
the  areas  of  the  rectangles  is  unity,  the  sum  of  the  areas  of  any 
number  of  adjacent  rectangles  is  equal  to  the  probability  that  a 
single  reading  falls  within  the  limits  represented  by  the  extremes 
of  the  bases  of  the  rectangles. 

In  a  large  number  of  cases  it  has  been  found  that  the  plotted 
points  representing  the  frequency  distribution  of  direct  observa- 
tions follow  the  curve  representing  the  equation 

y  =  ae-»***  (1) 

where  e  is  the  base  of  the  hyperbolic  logarithmic  system,  approxi- 
mately equal  to  2.7183,  a  and  h  are  positive  constants,  and 
x  the  variable  distance  from  the  vertical  line  of  symmetry,  i.e., 
distance  from  the  true  value.  Thus,  x  represents  the  error  of 
a  reading.  It  is  also  found  that  the  line  of  symmetry  corre- 
sponds to  the  arithmetic  mean  of  the  readings.  (See  Figs.  99 
and  100.  Also  Figs,  in  Methods  of  Least  Squares,  by  G.  C. 
Comstock.) 


§115] 


PROBABILITY 


223 


If  the  area  under  the  curve  is  taken  as  unity,  equation  (1) 
becomes 

tf=-/£«^V  (2) 


-12-11-10-9  -8  -7  -6  -5  -4  -3-2  -1 
A 


0    1 


234 
A' 


5    6    7    8   9  10  11  12 


FIG.  99. — Graphic  representation  of  frequency  distribution  of 
readings  given  in  Table  XIV.  The  circles  represent  the  distribution 
of  the  readings,  while  the  curve  is  the  probability  curve  calculated 
from  them.  The  equation  of  the  curve  is  y  =  Q.Q87e~0024x2.  The 
probable  error  of  the  mean  is  0.01  square  centimeters. 


FIG.  100. — Graphic  representation  of  frequency  distribution  of 
readings  of  the  elevation  of  a  bench  mark.  The  circles  represent  the 
distribution  of  the  readings,  while  the  curve  is  the  probability  curve 
calculated  from  them. 


In  what  follows  it  is  assumed  that  the  units  are  chosen  such  that 
the  area  under  the  curve  is  unity,  so  that  the  number  of  units  of 
area  bounded  by  the  curve  (2),  the  X-axis,  and  any  two  vertical 
lines  is  the  probability  that  a  single  reading  falls  within  the 
limits  corresponding  to  the  positions  of  the  vertical  lines,  or 


224  MATHEMATICS  [§115 

the  probability  that  a  single  reading  has  an  error  falling  within 
the  values  of  x  defining  the  vertical  lines.1 

The  curve  represented  by  equation  (2)  is  known  by  various 
names,  as  "the  probability  curve,"  "the  error  curve,"  "the  normal 
distribution  curve." 

While,  probably,  no  set  of  readings  which  the  student  will 
take,  or  will  have  occasion  to  study,  will  be  large  enough  in 
number  to  give  a  frequency  distribution  polygon  following  very 
closely  the  probability  curve,  yet  the  laws  and  formulas  derived 
from  this  law  will  be  used  and  applied  to  the  data  at  hand.  Thus, 
to  illustrate,  the  empirical  law  states  that  for  a  very,  very  large 
number  of  observations  the  line  of  symmetry  corresponds  to  the 
arithmetic  mean  of  the  readings,  and  hence  this  mean  gives  the 

1  For  students  who  have  a  knowledge  of  the  Calculus,  and  who  may  chance  to 
read  this  chapter,  a  few  proofs  of  statements,  which  must  of  necessity  be  taken  for 
granted  by  the  students  for  whom  this  book  is  intended,  are  inserted  in  footnotes. 

The  curve  for 

k-~* 
y  =  ae~  h  x 

is  symmetrical  with  respect  to  the  Y-axis  and  approaches  the  X-axis  asymptotically. 
y  has  a  maximum  value  when  x  =  0.  If  the  area  under  the  curve  is  unity,  it  follows 
that 


•J" 


, 

~h 


dx 


or,  if 


iru) 


a  =      /=-• 

vv 

Hence 


The  probability  that  a  single  reading  will  have  an  error  between  a  and  ft  is 

h 

'dx 


§116] 


PROBABILITY 


225 


true  value  of  the  measured  magnitude.  But  for  a  very  limited 
number  of  readings  the  arithmetic  mean  gives,  probably,  only  a 
very  close  approximation  to  the  true  value.  The  empirical 
law,  however,  does  seem  to  indicate  that  for  a  set  of  repeated 
readings  the  arithmetic  mean  is  the  best  mean  to  use  to  represent 
the  true  value. 

116.  Smoothing  the  Distributing  Polygon.  If  only  a  few  read- 
ings are  taken  the  frequency  distribution  polygon  will  in  general  be 
very  irregular.  A  more  regular  polygon  may  often  be  obtained  by 
grouping  the  readings  within  some  larger  interval.  This  method  is 
illustrated  in  column  III,  of  Table  XIV,  and  by  Fig.  101.  In  column 


6.3    6.6    6.9    7.2    7.5    7.8    8.1    8.4    8.7     9-0     9.3 

FIG.  101. — Frequency  distribution  polygon. 


Ill,  to  illustrate,  93  corresponding  to  area  127.5  is  the  sum  of  the 
frequencies  29, 33  and  31  of  column  II,  corresponding,  respectively, 
to  the  areas  127.4,  127.5  and  127.6.  In  Fig.  101  the  numbers  of 
column  III  are  plotted  against  areas. 

Another  method  of  smoothing  the  curve  is  to  consider  that  in 
a  larger  number  of  readings  the  frequency  of  any  one  would  be 
proportional  to  the  mean  of  the  observed  frequencies  of  it  and 
its  two  adjacent  readings.  The  frequencies  given  in  column  III 
of  Table  XIV  are  obtained  in  this  way  from  the  observed  frequen- 

15 


226  MATHEMATICS  [§117 

cies  given  in  column  II.1     In  Fig.  99  frequencies,  /,  are  plotted 
against  areas. 

117.  Measure  of  Precision;  Standard  Deviation.  For  a  set 
of  readings  it  may  not  be  enough  to  know  merely  the  value  of  the 
arithmetical  mean.  One  may  desire  to  know  what  degree  of 
confidence  may  be  placed  in  this  mean,  or  to  know  something  of 
the  agreement  or  disagreement  among  the  individual  readings. 
This  information  is  given  by  the  table  of  readings  or  by  the  fre- 
quency distribution  polygon,  but  it  is  desirable  to  have  some  definite 
mathematical  expression  giving  the  precision  of  the  set  of  readings. 

If  in  a  set  of  readings  the  number  of  small  errors  is  relatively 
large  compared  with  the  number  of  small  errors  of  a  second  set  of 
readings,  the  first  set  of  readings  would,  in  general,  command  more- 
confidence  than  the  second  set.  Of  the  two  corresponding  fre- 
quency distribution  polygons  the  first  would  be  tall  and  narrow  as 
compared  with  the  second.  Since  the  probability  curve  for  the 
first  set  of  readings  will  have  a  greater  maximum  ordinate  than 

h 

that  for  the  second  set,  and  since  —  j~-  is  the  value  of  the  maximum 

VV 

ordinate,  h  will  be  larger  for  the  first  set  than  for  the  second,  h 
is  called  the  measure  of  precision.  Hence  of  several  sets  of 
readings,  that  set  is  considered  the  best  for  which  h  is  a  maximum. 
Let  x  represent  the  distance  of  the  center  of  the  base  of  a 
rectangle  in  a  frequency  distribution  polygon  from  the  point  cor- 
responding to  the  arithmetic  mean.  In  Table  XIV,  these  values 
are  given  in  column  IV.  It  can  then  be  shown  that2 

A  _  2/«2  m 

2h*  ~       n 

1  In  this  case,  instead  of  taking  the  mean,  the  sum  of  the  three  readings  is  taken. 
This  is  done  to  avoid  the  division  by  3,  and  at  the  same  time  to  avoid  fractions. 
It  will  be  noticed  that  by  this  process  of  smoothing,  the  arithmetic  mean  is  not 
changed. 

2/i 

"  dv     (approximately) 


/""o 

I 

"Jo 


e~ldt,     where 


is  the  area  of  a  frequency  rectangle  (see  next  footnote).     If  the  horizontal 


§117]  PROBABILITY  227 

(where  n  represents  the  total  number  of  readings)  provided  that 
the  unit  of  v  is  the  interval  corresponding  to  the  base  of  the  frequency 
rectangle.1 


In  statistical  work  \/—  —  is  usually  denoted  by  <r  (sigma)  and  is 
called  the  standard  deviation.     Equation  (1)  may  then  be  written 

and  the  equation  of  the  probability  curve  may  be  written: 


In  practice,  instead  of  using  the  value  of  h  as  a  basis  of  comparison 
of  the  precision  of  sets  of  readings,  the  value  of  <r  is  used.  Since  a 
varies  inversely  as  h,  that  set  of  readings  is  considered  the  best 
which  gives  <r  the  smallest  value.2 

scale  is  chosen  so  that  At  is  unity,  the  sum  of  the  frequency  rectangles  will  be  unity 
if  V  =  f/n,  where  n  is  the  total  number  of  readings.  Thus: 

1 y/** 

2A2    ~  ^  n  • 

1  Frequency  rectangle,  as  here  used,  means  a  rectangle  of  the  frequency  distribu- 
tion polygon. 

2  From  equation  (1)  it  is  seen  that  for  h  to  have  a  maximum  value  S/t2  must  have 
a  minimum  value.     2/t>2  is  the  sum  of  the  squares  of  the  deviations  of  the  readings 
from  the  arithmetic  mean,  the  approximate  value,  or  2/c2  is  the  sum  of  the  squares 
of  the  residuals  as  they  are  called.     Thus  the  sum  of  the  squares  of  the  residuals 
must  be  a  minimum.     This  principle  is  sometimes  used  in  solving  simultaneous 
observation  equations    in    which  there    are    more    equation?   (corresponding    to  a 
large  number  of  readings)  than  unknown  numbers.     The  method  of  uniting  the 
equations  is  called  the  method  of  Least  Squares. 

An  observation  equation,  as  here  used,  is  an  equation  resulting  from  the  substitu- 
tion of  readings  in  an  equation  assumed,  or  known,  to  be  of  the  form  connecting  the 
measured  variables.  In  such  an  equation  coefficients  are  the  unknowns.  Thus, 
from  Table  VI,  if  x  represents  the  hydraulic  gradient  and  if  y  represents  velocity,  and 
if  it  is  assumed  that  the  law  connecting  y  and  x  is  linear,  i.e.,  y  =  ax  +  6,  there 
result,  by  substituting  the  measured  values  of  x  and  j/,  seven  equations,  viz.: 
16.9  =  31. 9o  +  6 

11.4  =     20.8o  +  6 

24.5  =     54.1o  +  6 
4.3  =     10. Oo  +  b 

10.1  =     21. Oo  +  6 

58.8  =  119.1o  +  b 

22.7  -     55.8o  +  6 

a  and  b  are  two  unknown  numbers  to  be  found,  having  given  seven  equations. 
The  problem  is  not  to  find  a  set  of  values  satisfying  any  two  equations,  but 
a  set  of  values  which  will  represent  the  most  probable  line  upon  which  the 
points  would  fall  if  there  were  no  errors  of  observation. 


228  MATHEMATICS  [§119 

118.  The  Probable  Error  of  a  Single  Reading.    The  probable 
error  of  a  single  reading,  Eg.r.,  is  a  number  such  that  if  it  be  sub- 
tracted from  and  then  added  to  the  arithmetic  mean,  the  proba- 
bility of  a  reading  falling  within  these  limits  is  exactly  equal  to 
1/2,  i.e.,  the  chance  of  a  reading  falling  within  these  limits  is 
exactly  equal  to  the  chance  of  its  falling  without.     Graphically, 
the  probable  error  determines  two  lines,  as  AB  and  A  'B',  Fig.  99, 
parallel  to  and  equidistant  from  the  line   of  symmetry  of  the 
probability  curve,  such  that  the  area  between  them,  the  curve, 
and  the  X-axis  is  one-half  unit.     (The  area  under  the  entire 
curve  and  the  X-axis  is  unity.) 

It  is  found  that1 

Es  r  =  0.6745  J—  =  0.6745  <r 

\    n 

For  the  data  given  in  Table  XIV, 

a  =  4.56 
Es.r.  =  3.08 
which  corresponds  to  0.308  square  centimeters. 

119.  The  Probable  Error  of  the  Arithmetic  Mean.     If  several 
sets  of  readings  of  the  same  magnitude  were  taken,  the  arithmetic 
means  of  these  sets  of  readings  would,  in  general,  disagree  among 
themselves.     Thus,  if  one  hundred  students  were  each  to  take 
an  equal  number  of  readings,  say  one  hundred,  of  the  area  bounded 
by  a  curve,  the  one  hundred  means  of  these  sets  would  probably 
not  all  agree.     From  these  one  hundred  means  a  frequency  dis- 
tribution polygon  could  be  constucted;  a  mean  (i.e.,  a  mean  of  the 
means)  and  a  probable  error  could  be  calculated.     This  probable 
error,  i.e.,  the  probable  error  of  a  single  mean  of  the  one  hundred 
means,  is  called  the  probable  error  of  the  mean  of  any  one  of  the 
sets  of  readings.     The  probable  error,  E,  of  the  mean  of  a  set  of 

i  The  value  of 


is  found  by  expanding  e  *  x  into  an  infinite  series  and  integrating  term  by  term 
Tables  for  different  values  of  the  upper  limit,  or  rather  for  x/<r,  or  hx,  are  usually 
given  in  treatises  on  the  Theory  of  Probability  and  in  treatises  on  Statistical 
Methods.  From  such  a  table  the  value  of  0.6745  for  x/a  is  found  to  correspond 
to  1/4. 


§120]  PROBABILITY  229 

readings,  bears  the  same  relation  to  a  set  of  means  as  the  probable 
error  of  a  single  reading  bears  to  the  readings  of  a  particular  set. 
A  formula  for  the  probable  error  of  the  mean  is 


The  probable  error  of  the  mean  of  the  readings  of  the  area  (Table 
XIV)  is 


., 

\/1068 

or  expressed  in  square  centimeters  E  =  0.01.  The  probable  error 
of  a  mean  is  usually  indicated  by  affixing  it  with  a  double  sign  to 
the  value  of  the  mean.  Thus,  in  the  above  expression, 

Area  =  127.66  ±  0.01 

This  does  not  mean  that  the  true  area  is  somewhere  between 
127.65  and  127.67,  but  that  with  the  data  at  hand  the  probability 
of  the  mean  of  a  second  set  of  the  same  number  of  readings 
falling  within  these  limits  is  one-half  (i.e.,  the  chances  are  one  to 
one  that  it  falls  within  these  limits). 

Exercises 

1.  Verify  all  numerical  work  done  in  connection  with  Table  XIV.1 

2.  Using  columns  I  and  II,  Table  XIV,  calculate  h,  <r,  Es.r.  and  E. 
Arrange  your  work  on  a  sheet  of  paper  form  M8  or  Ml 

120.  Biological  Measurements.  Frequently  distributions  of  bio- 
logical measurements  are  found  to  follow  to  a  marked  degree  the 
probability  law.  While  the  distribution  polygons  may  be  asym- 
metric, or  skew  (See  §  114),  the  formulas  for  standard  deviation, 
probable  error,  etc.,  are  still  applied  as  a  method  of  comparison.2 

Exercises 

1.  Construct  the  frequency  polygon,  find  the  mean,  the  standard 
deviation,  the  probable  error  of  the  mean,  for  the  readings  given  in 
Table  XV. 

1  In  treatises  on  the  Theory  of  Probability  and  in  treatises  on  Statistical  Methods 
are  usually  given  tables  of  squares. 

Crelle's  or  Peter's  Multiplication  Tables  or  the  smaller  table  by  Zimmermann 
may  also  be  found  very  helpful  in  numerical  calculations.  A  collection  of  tables 
and  formulas  is  given  in  Statistical  Methods  by  C.  B.  Davenport. 

'At  this  point  it  is  suggested  that  the  student  read  pages  419—441,  Principles  of 
Breeding,  by  E.  Davenport. 


230 


MATHEMATICS 


[§120 


TABLE  XV.— MEASUREMENTS  OF  THE  LENGTH  OF  327  EARS  OF  CORN' 


Length 
of  ear, 
inches 

Number  of 
ears,  or 
frequency 

Length 
of  ear, 
inches 

Number  of 
ears,  or 
frequency 

Length 
of  ear, 
inches 

Number  of 
ears,  or 
frequency 

V 

/ 

V 

/ 

t 

/ 

3.0 

1 

6.5 

8 

9.5 

63 

3.5 

0 

7.0 

12 

10.0 

38 

4.0 

1 

7.5 

19 

10.5 

21 

4.5 

0 

8.0 

32 

11.0 

8 

5.0 

2 

8.5 

40 

11.5 

2 

5.5 

3 

9.0 

67 

12.0 

1 

6.0 

9 

1  These  data  are  taken  from  Principles  of  Breeding,  page  421,  by  E.  Davenport. 


CHAPTER  X 
[SMALL  ERRORS] 

121.  Errors.  Whenever  an  attempt  is  made  to  measure  the 
magnitude  of  a  quantity  with  an  observing  instrument,  the  read- 
ings thus  obtained  do  not  represent  the  true  value  of  the  given 
magnitude.  There  is  a  difference  between  the  reading  of  the  in- 
strument, the  observation,  and  the  true  value  of  the  magnitude. 
This  difference  is  called  the  error  of  the  observation.  Errors 
of  observation  may  be  due  to  many  causes,  some  of  which  are: 
carelessness  on  the  part  of  the  observer  or  his  inability  to  handle 
the  measuring  instrument  properly;  inadjustment  of  the  measuring 
instrument;  limitations  due  to  the  least  count  of  the  instrument, 
etc.  By  the  least  count  of  the  measuring  instrument  is  meant 
the  smallest  part  of  a  unit  which  can  be  recorded  with  the  in- 
strument. To  illustrate:  when  a  volume  of  liquid  is  measured 
with  a  burette  graduated  to  cubic  centimeters  and  tenths  of 
cubic  centimeters,  the  readings  are  estimated  to  the  nearest 
tenth  of  one-tenth  of  a  cubic  centimeter;  the  least  count  of  the 
instrument  is  the  one  one-hundredth  part  of  one  cubic  centimeter. 

When  direct  measurements,  which  are  in  error,  are  used  for 
calculating  other  magnitudes,  these  calculated  magnitudes  will  in 
turn  be  in  error.  The  larger  the  errors  of  the  readings,  the 
larger  the  errors  of  the  calculated  quantities;  the  smaller  the 
errors  of  the  readings,  the  smaller  the  errors  of  the  calculated 
quantities.  When  the  errors  of  the  observations  are  known 
or  admitted,  it  is  at  times  desirable  to  know  the  corresponding 
error  in  the  calculated  magnitude  obtained  by  substituting 
the  readings  in  a  formula.  -If,  on  the  other  hand,  the  calculated 
magnitude  is  desired  only  to  a  certain  degree  of  accuracy,  it  is 
usually  desirable  to  know  the  degree  of  accuracy  to  which  the 
measurements  must  be  made. 

It  will- be  apparent  that  the  error  of  the  calculated  magnitude 
will  depend  upon  the  formula  in  which  the  readings  are  substituted. 

231 


232  MATHEMATICS  [§122 

Following  are  given  the  formulas  for  errors  due  to  the  simple 
algebraic  operations  and  then-  applications  to  some  of  the  formulas 
previously  derived  in  this  book. 

122.  Error  of  Product.  The  derivation  of  the  formula  for 
the  error  of  the  product  of  two  numbers  has  been  given  in  §  72, 
but  will  be  repeated  here. 

Let  a  and  b  be  two  readings.  Let  ci  and  62  be,  respectively, 
their  errors.  The  true  values  of  the  measured  magnitudes  are 
thus  assumed  to  be  a  ±  ei  and  b  ±  €2.  The  product  of  the 
two  magnitudes  is  then 

(a  +  €i)(6  +  €2)  =  ab  +  aez  +  bei  +  eiez. 

Since  ab  is  the  product  obtained  by  multiplying  the  two  readings 
together,  the  sum  of  the  last  three  terms  of  the  right-hand  side  of 
the  above  equation  is  the  error  of  the  product.  Since  e\  is  small 
compared  with  a  and  €2  is  small  compared  with  b,  eicz  is  very 
small  compared  with  ae2  +  bci  and  is  negligible.  Thus,  if 
ci  and  ez  are,  respectively,  the  errors  of  a  and  b,  ocz  +  &d 
is  the  possible  error  of  the  product  ab.  The  true  product  may 
fall  anywhere  between  ab  +  (ac2  +  &ci)  and  ab  —  (aez  +  bci). 
All  signs  in  the  error  are  taken  alike  in  order  to  obtain  the  greatest 
possible  error. 

In  what  follows,  expressions  of  the  form  a  =  212  +  0.3  will 
mean  a  reading  of  a  equal  to  the  first  number  (212),  and  a  possible 
error  equal  to  the  second  number  (0.3). 

If  a  =  212  +  0.3  and  b  =  116  ±  0.5,  the  error  in  the  product 
of  the  readings  is  aez  +  bci  =  (212)  (0.5)  +  (116)  (0.3)  =  150.8. 
Thus,  the  true  value  of  ab  is  between  24,592  +151  and 
24,592  —  151.  In  calculating  the  product  ab  by  multiplying  the 
two  readings  together  there  is  no  need  of  retaining  more  than 
the  first  three  digits,  and  the  product  should  be  written  24,600. 

Exercises 

1.  Show  that,  if  ei,  e2,  and  ei  are,  respectively,  the  errors  of  a,  b,  and 
c,  the  error  of  abc  is  a&es  +  bcei  +  ace2. 

Calculate  the  possible  error  of  the  product  for  each  of  the  following : 

2.  a  =  217     ±  0.2;  b  =  117     +  0.3. 

3.  a  =  1267  ±  0.5;  6  =  986    ±  0.4. 

4.  a  -  3176  ±  0.3;  b  =  1012  ±  0.3. 

5.  a  =  3176  ±  0.2;  6  =  1276  ±  0.3;  c  =  987  ±  0.4. 


§124]  SMALL  ERRORS  233 

123.  Error  of  Quotient.    Let  a  and  6  be  the  readings  with 
errors  of  ei   and  €2,   respectively.     The    quotient  of    the  true 
magnitudes  is 

a 

a  ±  *i  _  a         ~  b  2    '_^  _  a        ±ac2±bfi 
b  ±e2  =  b  H         6  ±  c2      ~  b  H "  ~b(b  ±  e2)  ' 

Since  -7  is  the  quotient  of  the  two  readings,  the  second  term 

of  the  right-hand  side  of  the  above  equation  is  the  error  of  the 
quotient.  Since  €2  is  small  compared  with  6,  the  quantity  b  +  e2 
may  be  replaced  by  &,  and  the  possible  error  of  the  quotient 
becomes 

ae2  +  tei 

If  a  =  317  +  0.5  and  6  =  110  +  0.6,  the  error  of  the  quotient 
a  . 

(317X0.6) +  (110)  (0.5)      ftn<? 
(HO)2 

Exercises 

Find  the  possible  error  of  the  quotient  for  each  of  the  following: 
l.o=  1127  ±  0.5;    b  =  56  ±  0.2. 

2.  a  =  2216  ±  0.6;    b  =  27  ±  0.1. 

3.  a  =  2716  +  0.3;    6  =  25  ±  0.5. 

124.  Error  of  the  Square.     Let  e  be  the  error  of  the  reading 
a.     The  square  of  the  true  value  is  (a  +  e)2  =  a2  +  2ae  +  e2. 
Since  a2  is  the  square  of  the  reading,  the  sum  of  the  last  two  terms 
of  the  right-hand  member  of  the  above  equation  is  the  error  of 
the    square.     The    possible    error,    2ac  +  e2,  may    be    written 
(2a  +  e)e,  and,  since  e  is  small  compared  with  2a,  the  expres- 
sion within  the  parentheses  reduces  to  2a  without   appreciable 
error.     The  possible  error  of  the  square  is  then 

2ae. 

If  a  =  112  ±  0.5,  the  error  of  a2  is  (2)  (112)  (0.5)  =  112. 
Since  a2  equals  12,544,  the  error  of  0.5  in  a,  or  about  1/2  of 
1  percent,  produces  an  error  of  about  1  percent  in  a2. 


234  MATHEMATICS  [§125 

Exercises 

Calculate  the  error  and  percent  error  of  the  square  for  each  of  the 
following  : 

1.  a  =  111  +  0.1. 

2.  a  =  316  +  0.2. 

3.  a  =  416  ±  0.6. 

4.  Show  that  the  percent  error  of  the  square  of  a  reading  is  the 
double  of  the  percent  error  of  the  reading. 

5.  Show  that  the  error  of  the  cube  of  a  +  e  is  3a2e. 

6.  Show  that  the  percent  error  of  the  cube  of  a  reading  is  the  treble 
of  the  percent  error  of  the  reading. 

7.  Show  that  the  error  of  the  fourth  power  of  a  +  e  is  4a3e. 

125.  Error  in  the  Square  Root.     Let  e  be  the   error  of  the 
reading  a.     The  square  root  of  the  true  value  is  then 

(a  ±  e)1^  E=  a*  ±  \a~\  +  \a~3\z  ±  ia-**e«  +     •    •    • 

Z  o  lO 

by  the  binomial   theorem.     (See  §  100.)     The   above   may   be 
written 


Since  e  is  small  compared  with  a,  the  terms  within  the  parenthe- 
ses, excepting  the  first,  are  small  compared  with  1,  and  may  be  ne- 
glected. Hence 

(a  ±  «)"  =  .«  ± 


approximately.     Since  a^  is  the   square   root   of    the    reading, 
f 
/==  is  the  error  of  the  square  root  of  a. 

If  a  =  121  +  0.5,  the  error  of  \/a  is 
0.5 


(2)  (11) 


=  0.023. 


Exercises 

Find  the  error  of  the  square  root  of  each  of  the  following  : 
l.o  =  121  ±  0.1. 

2.  a  =  146  +  0.3. 

3.  a  =  216  +  0.6. 


§126] 


SMALL  ERRORS 


235 


\c 


4.  Show  that  the  percent  error  of  the  square  root  of  a  reading  is 
one-half  of  the  percent  error  of  the  reading. 

6.  Show  that  the  error  of  the  cube  root  of  a  reading  is  s— sr' 

OCt'3 

6.  Show  that  the  percent  error  of  the  cube  root  of  a  reading  is  one- 
third  the  percent  error  of  the  reading. 

7.  Show  that  the  error  of  the  fourth  root  of  a  reading  a  is  T~57' 

126.  Error  of  Sine  a.  Let  e,  expressed  in  radians,  be  the 
error  of  the  measured  angle  a.  The  sine  of  the  true  angle  is  then 
sin  (a  ±  e).  Let  BOA,  Fig.  102,  be  the  angle  a  with  its  vertex 
at  the  center,  0,  of  a  circle 
having  a  unit  radius,  OA.  BE 
is  the  sine  of  the  reading  a,  CD 
is  the  sine  of  the  true  angle,  and 
CF  is  the  error  of  the  sine  due  to 
the  error  e  of  the  reading  a.  The 
angle  FCB  is  very  nearly  equal 
to  the  angle  a,  and  approaches 
it  as  e  approaches  zero.  The 
chord  CB  is  very  nearly  equal 
to  the  arc  CB.  FC  is  then  very 

nearly  equal  to  the  product  of  the  chord  CB  by  the  cosine  of 
FCB,  or  the  product  of  the  arc  CB  by  the  cosine  of  a.  But, 
since  e  is  expressed  in  radians,  the  arc  CB  equals  the  product 
of  e  by  the  radius  of  the  circle,  which  is  unity.  We  thus  have 
CF  =  e  cos  a,  approximately.  The  figure  is  drawn  for  e  posi- 
tive; the  same  conclusion  is  obtained  if  e  be  taken  negative. 
The  sign  of  e  regulates  the  sign  of  the  error  of  the  sine  of  a. 

To  illustrate    the    above    formula,    suppose   a  =  30°  ±15'. 
The  error  15'  expressed  in  radians  is 

Jg  4  -  0.00436. 

The  error  then  of  the  sine  a  is  (0.00436)  cos  30°, 
or  (0.00436)  (0.866)  =  0.0038. 

This  error  (or  the  error  of  any  trigonometric  function  of  any 
anclel  may  also  be  obtained  from  Table  XXII.  the  table  of 


FIG.  102. 


236  MATHEMATICS  [§127 

natural  trigonometric  functions.  The  sine  of  30°  is  0.5,  the  sine 
of  30°  +  15'  is  0.5038,  and  the  sine  of  30°  -  15'  is  0.4962.  In 
each  case  the  error  in  magnitude  is  0.0038,  the  same  as  found 
above  by  using  the  formula. 

Exercises 

1.  Find  by  substituting  in  the  formula  the  error  of  sin  a  for  each 
of  the  following: 

(a)  a  =  27°  ±  10'. 
(6)  a  =  87°  ±  20'. 

2.  Find  by  means  of  the  table  of  natural  trigonometric  functions  the 
error  of  sine  a  for  each  angle  given  in  exercise  1. 

3.  Find  the  error  of  cos  a,  if  a  =  26°  +  25'. 

4.  Find  the  error  of  tan  a,  if  a  =  62°  ±  17'. 

127.  Error  of  the  Area  of  a  Triangle  Computed  from  Two  Sides 
and  the  Included  Angle.  The  formula  for  the  area,  A,  of  a  tri- 
angle in  terms  of  two  sides,  a  and  b,  and  the  included  angle,  7,  is 

A  =  \  ab  sin  7. 

Let  ei  €2,  €3,  be  the  errors  of  a,  b,  and  7,  respectively.  Let  E 
represent  the  error  of  sin  7.  Then,  by  exercise  1,  §  122,  the 
error  of  A  is  \  (abE  +  feei  sin  7  +  aez  sin  7),1  where  E  —  63  cos  7. 
In  this  formula,  es  must  be  expressed  in  radians.  To  illustrate: 
let  a  =  111  +  0.2  rods;  b  =  72  ±  0.2  rods,  and  7  =  37°  +  15'. 
From  the  table  of  natural  functions,  E  is  found  to  be  0.0035. 
The  error  of  A  is  then 

i  [(111)  (72)  (0.0035)  +  (72)  (0.6018)  (0.2)  +  (111)  (0.6018)  (0.2)], 
or  50  square  rods. 

Exercises 

Find  the  error  of  the  area  of  the  triangle  for  each  of  the  following : 
l.o  =  126  ±  0.3  rods;  b  =  137  ±  0.3  rods;  y  =  47°  +  10'. 
2.  a  =  363  ±  0.8  rods;  b  =  216  +  0.5  rods;  7  =  26°  ±  15'. 

1  The  value  of  E  may  be  taken  from  a  table  of  natural  trigonometric  functions,  or 
the  expression  given  for  it  in  the  preceding  section  may  be  substituted,  in  which 
case  the  expression  for  the  error  of  A  becomes 

J(a&€scos-y  +  bei  sin  7  +  0*2  sin  7). 


§128]  SMALL  ERRORS  237 

128.  Error  of  the  Area  of  a  Triangle  Computed  from  Three  Sides. 

The  formula  for  the  area,  A,  of  a  triangle  in  terms  of  three  sides  is 


A  =  \/  s(s  —  a)(s  —  b)(s  —  c), 
where  s  =  |(a  +  b  +  c).     This  formula  may  be  put  in  the  form 


A  =  J      2o»6»  +  262c2  +  2  c2a2  -  a4  -  b*  -  c* 

Let  ei,  e2,  and  €3  be  the  errors  of  a,  b,  and  c,  respectively.  Then, 
by  §  124,  the  error  of  a2  is  2aei,  of  62  is  26e2,  of  c2  is  2ce3,  of 
a4  is  4a3ei,  of  64  is  4&3c2  and  of  c4  is  4c3e3.  By  §  122,  the  error 
of  a262  is  2a26e2  +  2ab2el}  of  62c2  is  26c2e2  +  262c3,  and  of  c2a2  is 
2ca2e3  +  2c2oei.  The  error  of  the  quantity  under  the  radical  sign 
is  then 


4o26e2  +  4a62d  +  462ce3  +  4&c2e2  +  4c2O€i  +  4ca2e3  - 


-  4c3c3, 

or 

4[a(62  +  c2 

-  a2)Cl  + 

7    /      0         1                9                    7   O\                  1 

0(0^  +  cr  —  6z)e2  -|- 

•  c(a2  + 

62  -  c2)63]. 

By  §  126  the 

error  of  A 

is  found  to  be 

1  Nf   4[a(62H 

-  c2  -  a2)e 

i  +  6(c2  +  a2  -  62)e2 

+  c(a2 

4-62-c2)e3], 

4  X 

2  V  2o262  - 

f  262c2  +   2c2a2  - 

a4  -64 

-c4 

or 

a(b2  +  c2 

-  a2)€l  + 

b(c2  +  a2  -  b2)€2  4 

-  c(a2  4 

•  b2  -  c2)«3 

84 

If  f\,  (2,  and  c3  are  each  k  percent  of  the  length  of  the  sides  of 
which  they  are  errors,  we  have 

ka 
61  =  i^ 

kb 

e2  =  - 
100 

^_ 

€3  ~  100 

Substituting  these  values  in  the  above  expression  for  the  error 
of  A,  it  reduces  to 

(b2  +  c2  -  a2)  +  6'(c*  +  a2  -  62)  +  c2(a2  +  62  -  c2)  ]  , 


238  MATHEMATICS  [§128 

or 

+  26V  +  2c2a2  -  a4  -  64  -  c4), 


or 


k  2kA 


or 


SWA™        zoo 

This  expression  says  that  if  the  same  percent  of  error,  k,  is  made 
in  measuring  the  three  sides  of  a  triangle,  the  error  of  the  calculated 
area  is  2k  percent.  It  is  to  be  noted  that  in  order  for  this  state- 
ment to  be  true  the  three  errors  must  have  the  same  algebraic  sign. 
To  illustrate:  if  a  =  100  ±0.1  foot,  b  =  200  ±  0.2  foot,  and 
c  =  150  +  0.1  foot,  the  posible  error  of  A  is 

525,000  +  300,000  +  412,500 

or  21  square  feet  or  a  possible  error  of  about  3/10  of  1  per- 
cent of  the  area. 

Exercises 

1.  Find  the  possible  error  of  the  area  of  the  triangle  if  a  =  127  ±  0.2 
rod,  and  b  =  263  +  0.4  rod,  and  c  =  290  ±  0.5  rod. 

2.  In  a  triangle  a  =  100  +  0.3,  0  =  37°  +  15',  and  7  =  52°  ±  15'. 
Calculate  the  possible  error  of  b. 

HINT:     Use  the  law  of  sines  and  note  that  the  possible  error  of  a 
is  the  sum  of  the  errors  of  /3  and  7. 

3.  If  the  error  of  a  is  e,  expressed  in  radians,  show  that  the  error 
of  cos  a  is  e  sin  a. 

4.  If  a  =  100  +  0.2,  b  =  200  +  0.3,  and  «  =  34°  +  15',  find  the 
possible  error  of 


COS    a 


CHAPTER  XI 
[POINT,  PLANE,  AND  LINE  IN  SPACE] 

129.  Rectangular  Space  Coordinates.  Through  any  point  0, 
Fig.  103,  are  drawn  three  right  lines,  X'X,  Y'Y,  and  Z'Z,  mutually 
at  right  angles.  The  point  0  is  called  the  origin  of  coordinates. 
The  three  lines  X'X,  Y'Y,  and  Z'Z  are  called,  respectively,  the 
X-axis,  the  F-axis,  and  the  Z-axis  of  coordinates.  The  three 
planes  determined  by  the  three  axes  taking  them  two  by  two  are 
called  the  coordinate  planes.  The  coordinate  planes  are  mutually 
at  right  angles.  The  plane  determined  by  the  X-  and  F-axes  is 
called  the  XY- plane,  by  the  F-  and  Z-axes  the  FZ-plane,  and 
by  the  X-  and  Z-axes  the  XZ-plane. 

The  three  coordinate  planes  divide  space  into  eight  parts 
called  octants.  The  octants  are  numbered  from  one  to  eight. 
The  first  octant  is  in  front  of  the  XY- plane,  to  the  right  of  the 
FZ-plane,  and  above  the  XZ-plane.  The  second  octant  is  back 
of  the  XF-plane,  to  the  right  of  the  YZ- plane,  and  above  the 
XZ-plane.  The  third  octant  is  back  of  the  XF-plane,  to  the  left 
of  the  FZ-plane,  and  above  the  XZ-plane.  The  fourth  octant 
is  in  front  of  the  XY- plane,  to  the  left  of  the  YZ- plane,  and 
above  the  XZ-plane.  The  fifth,  sixth,  seventh,  and  eighth 
octants  are  below  the  first,  second,  third  and  fourth  octants, 
respectively. 

Distances  measured  along,  or  parallel  to,  the  X-axis  to  the 
right  from  0  are  called  positive,  to  the  left  from  0  negative. 
Distances  measured  along,  or  parallel  to,  the  F-axis  up  from  0 
are  called  positive,  down  from  0  negative.  Distances  measured 
along,  or  parallel  to,  the  Z-axis  out  from  0  are  called  positive, 
back  from  0  negative. 

Any  point,  P,  in  space  is  located  by  giving  its  distance  from 
each  of  the  three  coordinate  planes.  FP,  or  OA,  Fig.  103,  its 
distance  from  the  FZ-plane,  is  called  its  x-coordinate;  DP,  or 
OC,  its  distance  from  the  XZ-plane,  is  called  its  y-coordinate; 

239 


240 


MATHEMATICS 


[§130 


and  EP,  or  OB,  its  distance  from  the  .XT-plane,  is  called  its 
z-coordinate.  The  algebraic  signs  of  the  three  coordinates  of 
points  in  the  eight  octants  are  given  in  the  following  table: 


Octant 

X 

y 

z 

First  

+ 

+ 

+ 

Second  

+ 

+ 

Third  

+ 

Fourth  

+ 

+ 

Fifth  

+ 

+ 

Sixth  

+ 

Seventh  

_ 

Eiehth.. 

4- 

Since  the  axes  of  coordinates  are  taken  mutually  at  right 
angles,  the  three  coordinates  of  a  point  P  are  called  rectangular 
coordinates.  In  designating  a  point  its  three  rectangular  co- 
ordinates are  written  in  the  order:  the  x-coordinate  first,  then 
the  ^/-coordinate,  and  then  the  z-coordinate.  Thus,  the  point 
(1,  2,  3)  is  in  the  first  octant,  one  unit  from  the  FZ-plane,  two 
units  from  theXZ-plane,  and  three  units  from  the  .XT-plane.  The 
point  ( —  1,  —  2,  3)  is  in  the  eighth  octant,  one  unit  from  the 
FZ-plane,  two  units  from  the  XZ-pl&ne,  and  three  units  from 
the  .XT-plane.  The  point  (  —  1,  —  1,  —  2)  is  in  the  seventh 
octant. 


3.  (-2,  -  1,  -5). 


Exercises 

Locate  the  following  points: 

1.  (1,  1,  1).  2.  (3,  -  2,  1). 

4.  (3,  -2,  -3).        5.  (-3,4,3). 

130.  Distance  between  Two  Points.  The  distance  of  P,  Fig. 
103,  from  the  origin  is  the  length  of  the  diagonal  of  the  rectangular 
parallelepiped  whose  edges  are  x,  y,  z,  the  coordinates  of  P. 
The  length  of  this  diagonal,  or  the  distance  of  P  from  the  origin 

is  A/a;2  +  i/2  +  22. 
Let  PI  and  Pz,  Fig.  104,  be  any  two  points  whose  coordinates 


§131] 


POINT,  PLANE,  AND  LINE  IN  SPACE 


241 


are,  respectively,  xi,  y\,  z\,  and  x2,  yz,  zz. 
that  the  distance  between  PI  and  Pz  is 


The  student  will  show 


131.  The  Locus  of  an  Equation.1  In  geometry  of  two  dimen- 
oxons  the  equation  x  =  2  represents  a  straight  line  parallel  to  the 
7-axis,  two  units  to  its  right.  (See  §  26.)  In  geometry  of 
three  dimensions  the  same  equation,  x  =  2,  represents  a  plane 


sions 


FIG.  104. 


parallel  to  the  7Z-plane,  two  units  to  its  right.  For,  the  only 
condition  to  be  satisfied  is  that  the  y-coordinate  of  the  point  is 
+  2;  the  coordinates  of  all  points  of  the  plane  satisfy  this  condition, 
and  the  coordinates  of  a  point  not  of  the  plane  do  not  satisfy 
the  condition.  The  locus  of  all  points  satisfying  the  condition 
x  =  2  is  a  plane  parallel  to  the  FZ-plane  two  units  to  its  right. 
The  locus  of  a  point  satisfying  the  condition  x  =  5  is  a  plane 
parallel  to  the  7Z-plane  five  units  to  its  right,  x  =  —  2  represents 
a  plane  parallel  to  the  FZ-plane  two  units  to  its  left,  x  =  0 
represents  the  FZ-plane.  x  =  k,  where  k  is  any  constant,  repre- 
sents a  plane  parallel  to  the  FZ-plane  k  units  from  it;  to  the  right 
if  k  is  positive,  to  the  left  if  k  is  negative.  Similarly,  y  =  k 
represents  a  plane  parallel  to  the  XZ-plane,  and  z  =  k  represents  a 
plane  parallel  to  the  XT-plane. 

1  The  locus  of  an  equation  is  the  locus  of  a  point  whose   coordinates,  x,  y,  and  c, 
satisfy  the  equation. 
16 


242  MATHEMATICS  [§131 

Any  equation  between  x,  y,  and  z,  when  solved  for  z,  takes  the 
form  z  =  f(x,  y) .  (The  right-hand  member  of  this  equation, 
f(x,  y),  is  read  "function  of  x  and  y,"  and  is  an  abbreviation  for 
the  right-hand  side  of  the  equation  when  solved  for  z.  The  f(x,  y) 
may  contain  x's  and  y's,  but  it  does  not  contain  z.)  If,  when  to  x 
and  y  in  the  equation  z  =  f(x,  y)  any  two  values  are  assigned, 
there  corresponds  a  real  value  of  z,  the  three  values,  one  for  x,  one 
for  y,  and  one  for  z,  locate  a  point  in  space  whose  coordinates 
satisfy  the  equation.  If  a  second  set  of  values  is  assigned  to  x 
and  y,  and  a  second,  or  the  same,  value  for  z  found,  a  second  point 
is  located  in  space.  If  the  difference  between  the  two  z-values 
and  the  difference  between  the  two  ^/-values  is  made  less  and  less, 
the  difference  between  the  two  z-values  will  in  general  grow  less 
and  less.  The  locus  of  a  point  whose  coordinates  satisfy  an  equa- 
tion is  then  seen  to  be  a  surface.  Thus,  in  the  equation 

z  =  ±  \/81  -  (z2  +  7/2) 

if  any  set  of  values  is  assigned  to  x  and  y  such  that  x2  +  y2  <  81, 
there  correspond  two  real  values  of  z,  numerically  equal  but  with 
opposite  signs.  If  a  second  set  of  values  is  assigned  to  x  and  y, 
a  second  (or  the  same)  pair  of  values  for  z  is  obtained.  If  the 
difference  between  the  two  x-values  and  the  difference  between 
the  two  ?/-values  are  made  smaller  and  smaller,  the  difference  be- 
tween the  two  positive  values  and  between  the  two  negative 
values  of  z  are  made  smaller  and  smaller.  As  x  and  y  are  made 
to  vary  continuously,  z  varies  continuously  and  the  equation  rep- 
resents a  surface  in  space.  If  the  values  assigned  to  x  and  y  are 
such  that  x2  +  y2  >  81,  the  corresponding  values  of  z  are  im- 
aginary, which  means  that  the  surface  does  not  extend  to  a 
point  whose  x  and  y  are  the  values  assigned  to  x  and  y  in  the 
equation.  Thus,  if  x  and  y  are  each  given  the  value  7  the  equa- 
tion reduces  to  z  =  ±  V  —  17,  an  imaginary.  This  says  that  the 
surface  represented  by  the  equation  contains  no  point  whose  x- 
and  y-coordinates  are  each  equal  to  7,  or  in  other  words,  a  line 
parallel  to  the  z-axis  seven  units  each  from  the  XZ-  and  YZ-co- 
ordinate  planes  does  not  pierce  nor  touch  the  surface. 

The  equation  z  =  ±  V  81  —  (z2  +  ?/2)  may  be  put  in  the  form 
x2  +  yz  +  z2  =  81.  The  expression  z2  +  y2  +  z2  is  the  square  of 


§132]  POINT,  PLANE,  AND  LINE  IN  SPACE  243 

the  distance  from  the  origin  to  the  point  whose  coordinates  are 
x,  y  and  z.  (See  §  130.)  This  expression  being  a  constant,  81, 
shows  that  all  points  whose  coordinates  satisfy  the  equation  must 
be  at  a  distance  9  from  the  origin.  The  locus  of  the  equation  is 
then  a  sphere  with  center  at  the  origin  and  radius  equal  to  9. 

132.  The  Equation  of  a  Locus.     The  equation  of  the  locus  of  a 
point  is  an  equation  satisfied  by  the  coordinates  of  all  points  of  the 
locus  and  by  the  coordinates  of  no  other  point.     Thus,  the  equa- 
tion of  a  sphere,  center  at  the  origin  and  radius  equal  to  10,  is 
x2  +  y*  +  z*  =  100.     The  equation  of  the  XZ-plane  is  y  =  0. 
The  equation  of  the  XY-p\a,ne  is  z  =  0.     The  equation  of  the 
YZ- plane  is  x  =  0.     The  equation  of  the  XY-  and  XZ-planes 
considered  as  a  single  locus  is  yz  =  0.     The  equation  of  the  three 
coordinate  planes  considered  as  a  single  locus  is  xyz  =  0. 

Exercises 

1.  Find  the  distance  between  the  following  pairs  of  points : 

(a)  (1,  1,  1)  and  (3,  2,  1).  (6)  (3,  -  2,  6)  and  (-  2,  1,  7). 

(c)   (-1,  -2,  -3)  and  (2,  6,  -1). 

2.  What  loci  are  represented  by  the  following  equations : 

(a)  x  =  3.  (6)      x  =  0.  (c)  y  =  -  6. 

(d)  y  =  0.  (e)      z  =  0.  (/)  z  =  -  10. 

(ff)  x*  +  t/2  +  z2  =  1.   (h)     x*  +y*  +z*  =  25. 

(i)  xy  =  0.  (7)  xyz  =  0.  (k)  x  =  y. 

(I)  x  =  z.  (TO)     y  =  z. 

3.  Find  the  equations  of  the  locus  of  (a)  a  point  ten  units  from  the 
origin;  (6)  a  point  ten  units  from  the  point  (3,  1,  2);  (c)  a  point  equi- 
distant from  the  XZ-  and  the  FZ-coordinate  planes. 

133.  The   Direction   Cosines   of   a  Line.    Let  P,   Fig.    103, 
be  any  point  on  the  line  OP  passing  through  the  origin.    The 
projections  of  OP  upon  the  three  coordinate  axes  are  the  co- 
ordinates of  the  point  P.    Let  a,  ft,  and  7  be  the  angles,  called 
the  direction  angles  of  the  line  OP,  between  OP  and  the  positive 
directions  of  the  x-,  y-,  and  z-axes,  respectively.     The  cosines  of 
these  angles  are  called  the  direction  cosines  of  the  line  OP. 


244 


MATHEMATICS 


[§134 


The  direction  angles  of  a  line  through  the  origin  are  positive  and 
do  not  exceed  180°.     From  the  figure  it  is  seen  that 
x  =  OP  cos  a 
y  =  OP  cos  (8 
z  =  OP  cos  7 
Squaring  and  adding  these  three  equations, 

z2  +  y2  +  z2  =  OP2(cos2  a  +  cos2  0  +  cos2  7). 
But,  since 

OP2  =  x2  +  ?/  +  z\ 
it  follows  that 

cos2  a  +  cos2/3  +  cos2  7  =  1.  (1) 

The  direction  angles  of  a  line  not  passing  through  the  origin  are 
defined  as  the  direction  angles  of  a  parallel  line  passing  through 
the  origin. 

The  sum  of  the  squares  of  the  direction  cosines  of  a  line  is  equal 
to  unity. 

134.  The  Projection  of  a  Broken  Line  upon  a  Given  Line. 
Let  ABODE,  Fig.  105,  be  any  broken  line  in  the  plane  or  in 


A' 


B'       a' 
FIG.  105. 


E' 


space.  From  the  points  A,  B,  C,  D,  and  E  perpendiculars  are 
dropped  upon  the  line  OP  meeting  it  at  the  points  A',  B',  C', 
D',  and  E'.  A'B',  B'C',  C'D',  and  D'E'  are,  respectively,  the  pro- 
jections of  AB,  BC,  CD,  and  DE.  Let  a,  p,  7,  and  5  be  the 
angles  between  the  directed  segments  of  AE  and  the  line  OP. 
A'B'  =  AB  cos  a,  B'C'  =  BC  cos  0,  C'D'  =  CD  cos  7  (which  is 
negative,  as  7  is  obtuse)  and  D'E'  =  DE  cos  6.  Then  A'E', 


§135] 


POINT,  PLANE,  AND  LINE  IN  SPACE 


245 


the  projection  of  a  directed  broken  line  upon  any  directed  line, 
is  the  sum  of  the  products  of  the  length  of  each  segment  of  the  broken 
line  by  the  cosine  of  the  angle  between  it  and  the  directed  line  upon 
which  it  is  projected. 

135.  The  Normal  Equation  of  a  Plane.  A  plane  is  determined 
if  its  normal  distance,  p,  from  the  origin,  and  the  direction  angles 
of  this  normal  are  known.  Let  ON,  or  p,  Fig.  106,  be  the  normal 


FIG.  106. 


drawn  from  the  origin  to  the  plane  ABC.  Let  a,  (3,  and  7  be 
the  direction  angles  of  this  normal.  Let  P  be  any  point  in  the 
plane.  Draw  PN,  which  is  perpendicular  to  ON.  The  sum  of 
the  projections  of  x,  y,  z,  and  PN  upon  the  normal  is  equal  to 
p.  The  projection  of  x  upon  the  normal  is  x  cos  a,  of  y  is  y.cos  @, 
and  of  z  is  z  cos  7.  The  projection  of  PN  upon  the  normal  is 
zero,  since  ONP  is  a  right  angle.  It  then  follows  that 

x  cos  a  +  y  cos  /3  +  z  cos  7  =  p,  (2) 

which  is  called  the  normal  equation  of  the  plane. 

In  the  equation  Ax  +  By  +  Cz  =  D,  let  A,  B,  C,  and  D  be  any 
real  constants.    The  equation  is  then  the  general  equation  of  the 


246  MATHEMATICS  [§135 


first  degree  in  x,  y,  and  z.    Upon  dividing  by  ±  V  A2  +  B*  +  C'2, 
this  equation  reduces  to 

=          ^x  +  -  B 


±  VA2  +  #2  +  c2    "  ±  VA2  +  £2  +  c2 


The  sign  before  the  radical  is  to  be  chosen  so  as  to  make  the 
right-hand  side  of  the  equation  positive. 

Since  the  numerical  values  of  the  coefficients  of  x,  y,  and  z 
in  equation  (3)  cannot  exceed  unity,  they  are  each  the  cosine  of 
some  angle.  Call  these  angles,  respectively,  a,  /3,  and  7.  Then 

cos2  «  +  cos2  ft  +  cos2  7  =  A2  +^2  +  c,  +  A.+%'+c. 

, 

h 


A2  +  52 

a,  /3  and  7  are  then,  by  §  133,  the  direction  cosines  of  some  line. 
The  left-hand  side  of  (3)  reduces  to 

x  cos  a  +  y  cos  /3  +  z  cos  7  , 

the  left-hand  side  of  the  normal  equation  of  a  plane;  and  since 
the  right-hand  side  of  (3)  is  a  positive  quantity,  call  it  p,  equation 
(3)  reduces  to 

x  cos  a  +  y  cos  j8  +  z  cos  7  =  p, 

the  normal  equation  of  a  plane. 


equation  of  a  plane  is  of  the  first  degree  in  x,  y,  and  z; 
and,  conversely,  every  equation  of  the  first  degree  in  x,  y,  and  z  is 
the  equation  of  a  plane.  To  reduce  Ax  +  By  +  Cz  =  D  to  the 
normal  form,  divide  the  equation  through  by  ±  V  A2  +  52  +  C2, 
choosing  that  sign  before  the  radical  which  mil  make  the  right-hand 
side  positive. 

To  illustrate:  let  x  +  2y  —  3z  =  —  2  be  the  given  equation. 
To  reduce  it  to  the  normal  form  divide  by  —  \S\4,  which  gives 

1232 


§136]  POINT,  PLANE,  AND  LINE  IN  SPACE  247 

2  1 

The  plane  is  — 1=  units  from  the  origin.    The  cos  a,  —  —j=.  is 

V14  VI* 

negative,  which  says  that  the  normal  drawn  from  the  origin  to  the 

plane  extends  into   the  third,  fourth,  seventh  or  eighth  octant. 

2 
The  cos  /3, .^   is  negative,  which  says  that  the  normal  from 

the  origin  to  the  plane  extends  into  the  fifth,  sixth,  seventh  or  eighth 

3 

octant.  The  cos  y,  +  ,— >  is  positive,  which  says  that  the  nor- 
mal drawn  from  the  origin  to  the  plane  extends  into  the  first, 
fourth,  fifth  or  eighth  octant.  The  three  direction  cosines,  then, 
show  that  the  normal  drawn  from  the  origin  to  the  plane  extends 
into  the  eighth  octant;  or  the  plane  and  the  three  coordinate 
planes  form  a  tetrahedron  in  the  eighth  octant. 


Exercises 

Reduce  each  of  the  following  equations  of  planes  to  the  normal 
form,  find  the  distance  of  each  plane  from  the  origin,  and  locate  the 
octant  in  which  the  plane  forms  a  tetrahedron  with  the  coordinate 
planes. 

l.z+j/+z  =  l.  2.  x  -  y  +  z  =  6. 

3.  3x  -  2y  +  5z  -  6  =  0.  4.  2x  +  7y  -  z  +  7  =  0. 

136.  The  Slope  of  a  Plane,  Given  the  Elevation  of  Three  Points. 

In  this  section  is  described  a  graphic  method  for  determining  the 
slope,  both  in  magnitude  and  direction,  of  a  plane  surface,  if  the 
relative  elevations  of  any  three  of  its  points  are  known.  A 
description  and  an  illustration  of  the  method  will  be  given  before 
the  proof. 

Let  A,  B,  and  C,  Fig.  107,  be  the  three  points  whose  elevations 
are  known.  Suppose  A  the  highest.  Compute  the  gradient  (in 
feet  per  100  feet,  in  feet  per  1000  feet,  in  feet  per  mile,  or  in  any 
other  desired  unit)  from  A  to  B  and  from  A  to  C.  From  A 
toward  B  draw  AB',  whose  length,  drawn  to  scale,  represents  the 
gradient  from  A  to  B.  From  A  toward  C  draw  AC',  whose  length, 
to  scale,  represents  the  gradient  from  A  to  C.  At  the  points  B' 
and  C'  draw  lines,  respectively,  perpendicular  to  AB  and  AC. 


248 


MATHEMATICS 


[§136 


These  perpendiculars  intersect  at  the  point  X.  Draw  AX,  which 
represents  in  direction  and  magnitude  the  slope  of  the  plane  ABC. 
To  illustrate:  suppose  B  and  C  are,  respectively,  2.26  and  3.20 
feet  lower  than  A.  Suppose  AB  =  838  feet,  and  AC  =  866  feet 
(measured  along  the  horizontal).  The  gradient  of  the  plane  ABC 
in  the  direction  AB  is  2.7  feet  per  1000  feet;  and  in  the  direction  AC, 
3.7  feet  per  1000  feet.  Lay  off  AB'  and  AC',  respectively  equal 


to  2.7  and  3.7  inches,  thus  making  1  inch  represent  a  slope  of  1  foot 
per  1000  feet.1  AX  is  3.79  inches  long,  which  shows  that  any  line 
on  the  plane  parallel  to  AX  has  a  gradient  of  3.79  feet  per  1000 
feet.  The  direction  AX  is  that  of  greatest  slope. 

If  A,  B  and  C  were  points  upon  the  plane  surface  of  an 
uneroded  sedimentary  rock,  the  length  of  AX  would  represent 
the  dip,  and  the  direction  perpendicular  to  AX  would  be  the 
strike  of  the  stratum. 

If  A,  B  and  C  represent  points  upon  the  ground-water  plane,  AX 
would  represent  the  direction  in  which  the  water  would  be  seeping. 

1  The  drawing  here  given  is  reduced.  The  student  should  redraw  it  to  scale, 
and  check  the  results  given  here. 


§137]  POINT,  PLANE,  AND  LINE  IN  SPACE  249 

To  lay  out  the  direction  of  a  line  having  a  given  slope,  say  1  foot 
per  1000  feet,  proceed  as  follows:  Upon  AX  as  diameter  describe 
a  circle.  With  A  as  center  and  with  a  radius  equal  to  1  inch 
describe  arcs  cutting  the  circle  at  the  points  M  and  N.  AM  and 
AN  are  then  the  directions  having  a  drop  of  1  foot  per  1000  feet. 

To  find  the  elevation  of  any  point  P,  draw  AP  intersecting  the 
circle  AB'C'X  at  the  point  K.  The  length  AK  gives  the  slope  of 
the  plane  in  the  direction  AP.  With  this  slope  and  the  distance 
AP  the  elevation  of  P  relative  to  A  may  be  calculated.  In  our 
specific  illustration,  AK  is  3.25  inches,  indicating  a  drop  of  3.25 
feet  per  1000  feet  in  the  direction  AP.  The  length  AP  is  2.18 
inches,  making  P  436  feet  distant  from  A.  (The  scale  of  ABC 
is  1  inch  to  200  feet.)  From  these  two  measurements  P  is  found 
to  be  1.42  feet  lower  than  A. 

Exercises 

1.  A,  B  and  C  are  three  points  on  a  plane.  AB  =  500  feet,  AC  = 
600  feet,  BC  =  400  feet,  measured  on  the  horizontal.  The  elevations 
of  the  plane  at  the  points  A,  B,  and  C  are,  respectively,  137.6  feet, 
132.4  feet,  and  135.7  feet.  Find  graphically  the  magnitude  and 
direction  of  the  slope  of  the  plane  ABC.  Find  the  elevations  of  two 
points  each  300  feet  from  B  and  200  feet  from  C.  Find  two  directions 
having  a  slope  of  3  feet  per  1000  feet.  Find  the  slope  in  the  directions 
from  A  to  two  points  each  300  feet  from  A  and  400  feet  from  B. 

[137.]  Proof  of  the  Construction  Given  in  the  Preceding  Section. 
In  the  proof  for  the  above  construction  let  A,  Fig.  108,  be 
taken  as  the  origin  of  coordinates,  and  let  the  XY- plane  pass 
through  the  point  B.  The  coordinates  of  A  are  then  0,  0,  and  0. 
Let  the  coordinates  of  B  be  a,  —  b,  and  0;  and  of  C  be  a',  —  b', 
and  c'.  The  general  equation  of  a  plane,  by  §  136,  is 

Ax  +  By  +  Cz  =  D.  (4) 

Since  the  plane  passes  through  the  origin,  (0,  0,  0)  must  satisfy 
its  equation,  which  shows  that  D  is  zero.  Since  the  plane  passes 
through  the  point  B,  its  coordinates  must  satisfy  the  equation, 
thus  giving 

aA-bB  =  0.  (5) 


250 


MATHEMATICS 


[§137 


Similarly,  the  coordinates  of  the  point  C  must  satisfy  the  equation, 
giving 

a' A  -  b'B  +  c'C  =  0.  (6) 

Solving  equations  (5)  and  (6)  for  A  and  B  we  find 

bc'C 


C.  (a,-b,  c) 
FlG.    108. 

With  these  values,  equation  (4)  reduces  to 

bc'x  +  ac'y  +  (ab1  -  a'b)z  =  0.  (7) 

The  direction  cosines  of  the  upward  drawn  normal  to  this  plane 
are,  by  §  133, 

be'       ac'          .  ab'  —  a'b 
—T=>    — T=>   and  —  — j=~ 

Vi   Vi  Vi 

where 

/  =  6V2  +  aV2  +  (ab'-  a'6)2. 

6  b' 

The  gradient  of  the  line  AB  is    — ,  and  of  A  C  is      /  .,   ,    =j=- 

a  v  a  *•  -f-  c 

The  equations  of  the  lines  B'X  and  C'X,  Fig.  107,  on  the  XZ- 
plane  are,  respectively,  x  =  -  and    a'x  +  c'z  =  b'.      The    co- 

(Z 

ordinates  of  the  point  X   are  then  x  =  —  andz  =  -   —, 


§137]  POINT,  PLANE,  AND  LINE  IN  SPACE  251 

Let  AT,  Fig.  108,  be  the  upward  drawn  normal  to  the  plane. 
Take    T   such   that   AT  =  -y^-,.    The   coordinates  of  S,   the 

.AC 

projection    of     T    upon     the     .X"Z-plane,     are     —,  --  —   and 

ac 


V7  ab'  -  a'b  &  ab'  -  a'b      ,  .  , 

—  r       —  7^  —  '   or,  —  and  -    —  ;  —  ,  which  are  seen  to  be  the 

ac'        \/I  '  a  ac' 


coordinates  of  the  point  X.  This  shows  that  the  slope  of  the 
plane  is  in  the  direction  AX.  We  shall  now  show  that  the  length 
of  AX  represents  the  magnitude  of  the  slope  of  the  plane. 

The  distance  AX  is  ~  Vb*c'z+(ab'-a'b)*.     By   substituting 

etc 

the  coordinates  of  X  in  the  equation  of  the  plane  passing  through 
the  points  A,  B,  and  C,  we  find  the  plane  to  be 


units  directly  beneath  X.     Upon  dividing  this  expression  by  the 
length  of  AX  we  have 


—  V&V2  +  (ab'  -  a'b)2, 

which  is  equal  to  the  length  of  AX  found  above.     This  shows  that 
AX  represents  the  magnitude  of  the  slope  of  the  plane  ABC. 

Since  any  angle  inscribed  within  a  semicircle  is  a  right  angle, 
a  perpendicular  drawn  to  AK,  Fig.  107,  at  the  point  K  will  pass 
through  X.  From  what  was  proved  above,  AK  must  represent 
the  gradient  of  the  plane  ABC  in  the  direction  AP.  For  the  same 
reason,  AM  represents  the  slope  (given)  in  the  direction  AM. 


CHAPTER  XII 
[MAXIMA  AND  MINIMA] 

138.  Maxima  and  Minima  Defined.  Let  y  be  a  function  of 
x  (See  §  45).  If,  as  x  increases,  y  increases  to  a  certain  value, 
M,  and  then  begins  to  decrease,  M  is  called  a  maximum  value 
of  y.  If,  as  x  increases,  y  decreases  to  a  certain  value,  m,  and 
then  begins  to  increase,  ra  is  called  a  minimum  value  of  y. 

A  maximum  value  of  a  function  is  not  necessarily  the  largest 
value  the  function  may  have;  and  a  minimum  value  is  not  neces- 
sarily its  smallest  value.  A  function  may  have  more  than  one 
maximum,  and  more  than  one  minimum,  and  a  function  may  have 
neither  maxima  nor  minima. 


£>' 


FIG.  109. 


Graphically  the  maxima  and  minima  values  of  a  function  are, 
respectively,  the  lengths  of  the  ordinates  of  the  high  and  low 
points  of  the  graph  of  the  function. 

Let  ABODE,  Fig.  109,  be  the  graph  of  the  function  y  =  /(x). 
As  x  increases  from  zero  to  OA',  y  increases  to  the  value  repre- 
sented by  the  ordinate  A' A.  As  x  increases  from  OA'  to  OB', 
y  decreases.  Then  by  definition,  A'A  represents  a  maximum 
value  of  the  function.  Similarly  C'C  represents  a  second  maxi- 

252 


§139] 


MAXIMA  AND  MINIMA 


253 


mum  value,  and  B'B  and  D'D  represent  two  minimum  values. 
The  minimum  value  DD'  is  larger  than  the  maximum  value  A' A. 
139.  Maximum  and  Minimum  Values  by  Plotting.  When 
observed  data  connecting  two  variables  are  plotted  upon  rec- 
tangular coordinate  paper,  and  a  smooth  curve  is  drawn  through 
the  plotted  points,  the  maximum  and  minimum  values  may  be 
found  approximately  from  the  graph.  Thus,  from  Fig.  15. 
the  maximum  air  temperature  was  approximately  67.5°  at  3  : 45, 


13^'-- 


«*- 

•<  —    --(13-2  *")  > 

FIG.  110. 

If  the  equation  of  the  curve  be  known  it  may  be  plotted  point  by 
point  and  the  approximate  values  of  the  maximum  and  minimum 
values  taken  from  the  graph.  To  illustrate  this  method  take  the 
following  problem:  From  a  square  piece  of  tin  13  inches  on  a 
side  an  equal  square  is  cut  from  each  corner.  The  edges  are  then 
turned  lip  forming  a  box.  Find  the  size  of  the  square  cut  from 
each  corner  such  that  the  volume  of  the  box  shall  be  a  maximum. 

Let  y  be  the  volume  of  the  box.  Let  x  be  the  side  of  the  square 
cut  out  from  each  corner,  Fig.  110.  The  volume  of  the  box  is 
then  y  =  (13  —  2x)2x.  By  assigning  to  x  values  and  calculat- 
ing the  corresponding  values  of  y  the  following  table  is  con- 
structed: 


254 


MATHEMATICS 


[§139 


X 

2/    ||    z 

y 

0 

0 

7/2 

126 

1/2 

72 

8/2 

100 

2/2 

121 

9/2 

72 

3/2 

150 

10/2 

45 

4/2 

162 

11/2 

22 

5/2 

160 

12/2 

6 

6/2 

147 

13/2 

0 

From  this  table  it  is  seen  that  a  maximum  value  of  y  exists 
when  x  lies  somewhere  between  f  and  f .  By  giving  to  x,  values 
nearer  together,  the  following  table  is  constructed: 


X 

V 

x 

y 

1.6 

153.7 

2.6 

158.2 

1.8 

159.0 

2.8 

153.3 

2.0 

162.0 

3.0 

147.0 

2.2 

162.7 

3.2 

139.4 

2.4 

161.4 

3.4 

130.7 

From  this  table  it  is  seen  that  the  maximum  value  of  y  occurs 
for  some  value  of  x  between  2.0  and  2.4    In  Fig.  Ill  is  given  the 


160 


150 


1.6 


1.8 


2.0 
X 

FIG.  111. 


2.2 


2.4 


2.6 


plot  of  the  function  for  values  of  x  ranging  from  f  to  f-.  From 
this  graph  it  is  seen  that  an  approximate  maximum  value  of  y 
is  162.7  A  closer  approximation  may  be  obtained  by  assigning 
to  x  a  larger  number  of  values  between  2.0  and  2.4. 


§140] 


MAXIMA  AND  MINIMA 


255 


Exercises 

1.  Same  as  the  illustrative  problem  above,  but  with  the  side  of  the 
square  of  tin  equal  to,  (a)  10  inches;  (6)  15  inches;  (c)  17  inches;  (d) 
20  inches. 

2.  If  a  represents  the  side  of  the  square  of  tin  in  exercise  1,  and  x 
the  side  of  the  square  cut  out,  calculate  the  ratio  x/a  for  each  case. 

3.  Find  the  dimensions  of  the  greatest  rectangle  that  can  be  in- 
scribed in  a  circle  with  radius  10. 

140.  The  function  ax2  +  &x  +  c.     In  §  31  it  was  shown  that 
the  function  axz  +  bx  +  c  has  a  maximum  or  a  minimum  value. 


-i 


-2 


(••*.-*) 

FIG.  112. 


Thus,  to  illustrate,  the  function  y  =  4x2  +  8x  —  2  may  be  put  in 
the  form 

y  =  4z2  +  Sx  +  1  -  2, 
or 

y  =  (2x  +  I)2  -  2, 
or 

7/  =  4(x  +  l/2)2-2. 

The  last  equation  shows  that  the  vertex  of  the  parabola  is  at  the 
point  ( —  TJ-,  —  2) ,  and  that  the  axis  of  the  parabola  extends  up 
from  the  vertex,  as  is  represented  in  Fig.  1 12.  The  function  then 


256  MATHEMATICS  [§141 

has  a  minimum  value  —  2,  which  is  the  value  corresponding  to 
value  x  =  —  ^-. 

Exercises 

Find  the  maximum  or  minimum  value  for  each  of  the  following 
functions  : 

1.  x2  +2x  +  1.  4.  9x2  +  3x  -  2.          7.  2x  -  x2. 

2.  x2  +  3x  -  6.  5.  7x2  _  2x  -  6.          8.  3  -  5x  -  2x2. 

3.  4x2  +  16x.  6.  8x2  -  3x  -  7.          9.  3  +  5x  -  2x2. 

10.  If  a  body  is  thrown  vertically  upward  with  an  initial  speed  of  a 
feet  per  second,  its  height,  h,  in  feet,  at  the  end  of  t  seconds  is  given 
by  the  equation  In  =  at  —  16.  li2.  To  what  height  will  the  body  rise 
if  thrown  with  an  initial  speed  of  32.2  feet  per  second?  When  will  it 
reach  this  height? 

141.  Maxima  and  Minima  Found  by  Limits.  In  some  problems 
the  maxima  or  minima  values  of  a  function  may  be  found  by  a 
method  illustrated  by  the  following  examples  : 

EXAMPLE  1  :  Find  the  maximum  and  minimum  values,  if  any, 
of  the  f  uncton 


2*4-1 

To  find  the  maximum  and  minimum  values  of  the  function  is 
to  find  the  ordinates  of  the  turning  points  of  the  graph  of  the 
equation 

a2  +  6 


Upon  solving  for  x  the  equation 


. 


x  =  y  ±V(y-2)(y  +  3)  (2) 

is  obtained. 

If,  in  equation  (2),  values  are  assigned  to  y  (y  =  k,  where  k 
is  some  constant)  corresponding  values  of  x  are  found;  thus 
locating  points  upon  the  graph  of  the  function.  These  values 
of  x  are  given  by 


x  =  k  ±  V(k  -  2)(fc  +  3).  (3) 

The  equation  y  =  k  represents  a  straight  line  parallel  to  the 
x-axis,  k  units  distant,  and  the  values  of  x  found  from  equation 


§141] 


MAXIMA  AND  MINIMA 


257 


(3)  are  the  z-coordinates  of  the  points  of  intersection  of  this 
straight  line  with  the  graph  of  equation  (2). 

The  second  factor,  k  +  3,  under  the  radical  of  equation  (3) 
is  positive  for  all  positive  values  of  k,  and  for  all  negative  values 
of  k  numerically  less  than  3,  or  positive  for  all  values  of  k  greater 
than  —  3.  It  is  zero  when  k  equals  —  3,  and  negative  when 
k  is  less  than  —  3.  The  first  factor  is  positive  when  k  is  greater 
than  2,  and  zero  when  k  equals  2,  and  negative  when  k  is  less 
than  2. 

When  both  factors  are  positive,  i.e.,  when  k  is  greater  than  2 
or  less  than  —  3,  x  is  real,  and  the  line  y  =  k  cuts  the  graph  of 
equation  (2)  in  two  real  points.  The  z-coordinates  of  these 
points  are  the  values  of  x  found  from  equation  (3).1  When  k 


FIG.  113. 

is  greater  than  —  3  and  less  than  2,  x  is  imaginary,  for  then  the 
quantity  under  the  radical  is  negative.  This  shows  that  all 
lines  of  the  system  y  =  k  between  the  lines  y  =  2  and  y  =  —  3, 
fail  to  cut  the  curve.  When  k  =  —  3  or  2  there  is  in  each  case 
but  one  value  (or  two  equal  values)  of  x.  The  lines  y  =  2  and 
y  =  —  3  are  tangent  to  the  curve. 

From  the  above  discussion  it  is  seen  that  the  graph  of  equa- 
tion (2)  consists  of  two  branches,  one  above  and  tangent  to  the 

1  The  student  should  plot  the  graph  of  equation  (2).     Give  to  y  in  (2),  or  to  k  in 
(3),  values  ranging  from  —  10  to  +  10. 
17 


258  MATHEMATICS  l§141 

line  y  =  2,  the  other  below  and  tangent  to  the  line  y  =  —  3. 
These  two  branches  extend  to  infinity,  one  in  the  positive,  the 
other  in  the  negative  y  direction,  for  x  is  real  for  all  positive  k 
values  greater  than  2,  and  for  all  negative  k  values  less  than  —  3. 
The  graph  has  two  turning  points,  one  with  a  ^-coordinate  2, 
the  other  with  a  7/-coordinate  —  3.  The  first  is  a  minimum  value, 
the  second  a  maximum  value  of  the  given  function.  The  mini- 
mum value  corresponds  to  the  value  2  for  x,  the  maximum  value 
corresponds  to  the  value  —  3  for  x. 

EXAMPLE   2:     Find  the  maximum  rectangle  that  can  be  in- 
scribed in  a  circle  with  radius  10. 

In  Fig.  113,  let  x  be  one-half  the  length  of  the  base,  and  z  one- 
half  the  length  of  the  altitude.     The  area,  y,  is 

y  =.4z«. (1) 

z  =  \/100  -  z2.  (2) 

Substituting  in  equation  (1) 

y  =  4z\/100  -  x*.  (3) 

Squaring  equation  (3) 

?/2  =  16z2(100  -  z2).  (4) 

Solving  (4)  for  xz 

X2  =  800+  V(200-y)(200  +  y)         (fi) 

16 

In  equation  (5)  the  quantity  under  the  radical  is  positive  if  y 
is  less  than  200  and  greater  than  -  200,  (-  200  <  y  <  200), 
and  z2  is  real.  Beyond  these  values  for  y,  x2  becomes  imaginary. 
200  then  is  the  maximum  value  of  y.  When  y  equals  200,  x2 
equals  50,  or  x  =  \/50,  and  from  equation  (2),  z  =  \/50.  The 
maximum  rectangle  is  then  a  square  with  a  side  equal  to  2 -x/50. 

Exercises 

1.  A  gutter  with  rectangular  cross  section  is  made  from  a  strip  of 
tin  30  inches  wide.     Find  the  width  of  the  portion  of  tin  bent  up  upon 
each  side  so  that  the  carrying  capacity  of  the  gutter  shall  be  a 
maximum. 

HINT  :    Let  y  be  the  area  of  the  cross  section,  and  let  x  be  the  width 
of  the  tin  bent  up.     Then  y  =  x(3Q  —  2x). 

2.  Find  the  length  of  the  lever,  exercise  6,  §  91,  which  will  make 
the  upward  pull  a  minimum. 


§142] 


MAXIMA  AND  MINIMA 


259 


142.  Another  Method  of  Finding  Maxima  and  Minima.    The 

method  considered  here  of  finding  the  maxima  and  minima  of  a 
function  depends  upon  a  study  of  the  slope  of  the  tangent  line  drawn 
to  the  curve  representing  the  function.  In  Fig.  109,  a  tangent  line 
drawn  to  the  curve  at  any  point  between  P  and  A  makes  an  acute 
angle  with  the  positive  direction  of  the  axis  of  x.  The  slope  of  any 
such  tangent  line  is  then  positive.  Any  tangent  line  drawn  to 
the  curve  between  the  points  A  and  B  makes  an  obtuse  angle  with 
the  positive  direction  of  the  axis  of  x.  The  slope  of  any  such  tan- 
gent line  is  then  negative.  The  slope  of  any  tangent  lines  drawn 
to  the  curve  between  the  points  B  and  C,  and  between  the  points 
D  and  E,  is  positive.  The  slope  of  any  tangent  lines  drawn  to  the 
curve  between  the  points  C  and  D  is  negative.  When  the  slope  of 
the  tangent  line  is  positive,  the  curve  ascends;  when  it  is  negative 


FIG.  114. 

the  curve  descends.  When  the  slope  of  the  tangent  line  changes 
from  positive  to  negative  the  curve  changes  from  ascending  to 
descending  and  possesses  a  high  point.  When  the  slope  of  the 
tangent  line  changes  from  negative  to  positive,  the  curve  changes 
from  descending  to  ascending  and  possesses  a  low  point. 

EXAMPLE  1 :     In  §  139  the  volume,  y,  of  a  box  formed  from  a 
square  piece  of  tin  13  inches  on  a  side  was  found  to  be 

y  =  (13  -  2z)2z  =  169x  -  52x2  +  4z3.  (1) 

Let  us  suppose  that  the  curve  in  fig.  1 14  represents  the  graph  of 
the  function.  Let  P  be  any  point  on  the  curve  whose  coordinates 


260  MATHEMATICS  [§142 

are  x  and  y.  Let  Q  be  any  other  point  on  the  curve  whose  co- 
ordinates are  x  +  h  and  y  +  k.  Since  these  points  are  on  the 
curve,  their  coordinates  must  satisfy  equation  (1),  from  which 
we  obtain 

y  =  169x  -  52x2  +  4x3  (2) 

and 

y  +  k  =  169(x  +  h)  -  52(x  +  h)2  +  4(x  +  ft)3.  (3) 

Equation  (3)  reduces  to 
y  +  k  =  169x  +  169ft  -  52x2  -  104ftx  -  52ft2  +  4x3 

+  12x2ft  +  12xft2  +  4ft3.      (4) 

Subtracting  equation  (2)  from  equation  (4)  there  results 
k  =  169ft  -  104ftx  +  52ft2  +  12ftx2  +  12xft2  +  4ft3, 
or 

|  =  169  -  104z  +  52ft  +  12.x2  +  12xft  +  4ft2.  (6) 

k/h  represents  the  slope  of  the  secant  line  drawn  through  the 
points  P  and  Q.  If  the  point  Q  is  taken  nearer  and  nearer  the  point 
P,  the  secant  line  rotates  about  the  point  P  and  approaches  the 
tangent  line  drawn  to  the  curve  at  that  point.  The  slope  of  the 
tangent  line  is  then  the  limit  of  the  slope  of  the  secant  line  as  the 
point  Q  approaches  the  point  P.  But,  as  Q  approaches  P,  h 
approaches  zero,  and  the  right-hand  side  of  equation  (6)  ap- 
proaches 169  —  1042  +  12x2.  Therefore,  the  slope  of  the  tangent 
line  drawn  to  the  curve  representing  the  volume  of  the  box  is 

r          1 04.          1  fiQ 
12x2-  104x  +  169,  or  12  I  x2  ~   jrf  *  +  ^ 

or 

slope  =12(x-f)(x-f).  (7) 

Since  the  side  of  the  square  of  tin  is  13  inches,  x,  the  side  of  the 
square  cut  from  each  corner,  must  be  less  than  6j  inches.     Thus, 

/        13\ 
the  last  factor,  ( x  — ^-j  ,  of  equation  (7)  is  always  negative.     The 

second  factor,  (x  —  -g-j,  is  negative  if  x  is  less  than  -^-,  and 

13 

positive  if  x  is  greater  than^--     The  expression  representing  the 

13 

slope  is  then  positive  for  values  of  x  ranging  from  0  to  -       and 


§142]  MAXIMA  AND  MINIMA  261 

for  these  values  of  x  the  curve  representing  the  function  rises. 

13 

For  all  values  of  x  ranging  from  -~-  to  13  the  slope  is  negative, 

and  the  curve  descends.     Thus,  the  function  representing  the 

13 

volume  of  the  box  has  a  maximum  value  when  x  =  -^  •    To  find 

o 

the  maximum  volume  substitute  this  value  for  x  in  equation  (1), 
which  then  gives 

y  =  162.74. 

EXAMPLE  2:    The  area  of  a  rectangle  inscribed  in  a  circle, 
radius  10,  is  given  by  equation  (3),  example  2,  §  141. 

y  =  4x\/100  -  x2,  (8) 

or 

y2  =  16x2(100  -  a;2) 

=  IGOOx2  -  16x4.  (9) 

A  value  of  x  which  makes  y  a  maximum  will  at  the  same  time  make 
y2  a  maximum.  For  brevity  call  y1  =  u,  and  equation  (9) 
becomes 

u  =  1600x2  -  16x4.  (10) 

As  in  the  preceding  example,  let  P  and  Q  be  any  two  points  upon 
the  curve,  Fig.  114,  which  represents  the  relation  between  u  and  x. 
If  x  and  u  are  the  coordinates  of  the  point  P,  and  x  +  h  and  u  +  k 
those  of  the  point  Q, 

u  =  IGOOx2  -  16x4  (11) 

and 
u  +  k  =  1600(x  +  hY  -  16(x  +  hY 

=  IGOOx2  +  3200/ix  +  IGOO/i2  -  16x4  -  64/ix3 

-  96/i2x2  -  64^3x  -  16/i4.     (12) 

Upon  subtracting  equation  (11)  from  equation  (12)  there  results 
k  =  3200Ax  +  1600/12  -  64/ix3  -  96/i2x2  -  64h3x  -  I6h*,  (13) 
or 

kh  =  3200*  +  1600A  -  64z3  -  96/ix2  -  64^2x  -  16A3.     (14) 

As  h  approaches  zero,  the  right-hand  side  of  equation  (14)  ap- 
proaches 3200x  —  64x3,  or 

slope  =  64z(50  -  x2)  =  64x( \/50  +  x)( \/50  -  x).      (15) 


262  MATHEMATICS  [§142 

From  the  right-hand  side  of  equation  (15)  it  is  seen  that  the  slope 
is  negative  if  £  is  greater  than  \/5Q,  and  positive  if  x  is  positive 
and  less  than  -y/50-  The  curve  ascends  for  all  values  of  x  from 
0  to  \/50,  and  descends  for  all  values  of  x  greater  than  \/50- 
Thus,  u  has  a  maximum  value  when  x  =  \/50-  When  u  is  a 
maximum,  y  is  a  maximum;  therefore  the  maximum  rectangle  is 
that  for  which  x  =  \/50.  The  remaining  discussion  of  this 
problem  has  been  given  in  §  141. 

EXAMPLE  3  :  Find  the  most  economical  proportion  of  a  closed 
cylindrical  can.  Let  z  be  the  altitude,  x  the  radius  of  the  base, 
v  the  volume,  and  y  the  surface  of  the  can.  The  problem  is  to  find 
the  ratio  x/z  which  will  make  y  a  minimum  for  any  constant 
value  of  v. 

V    =  TTZX2  (16) 

y  =  2irzx  +  2irx*  (17) 

Upon  eliminating  z  between  equations  (16)  and  (17) 

y  =  ~  +  2irx*.  (18) 

U/ 

2v 
y  +  k  =  —      +  27r(a;  +  h}\  (19) 


Subtracting  equation  (18)  from  equation  (19) 


+ 

-  2vh 


or  K 

or 

k 

h 

As  h  approaches  zero,  the  right-hand  side  of  equation  (20)  ap- 

2v 
preaches  4irx 2-     Then 

2v 

slope  =  lirx -j 

x2 

or 

A.™-    /  11  \ 

(21) 


§142]  MAXIMA  AND  MINIMA  263 

3   /  V 

If  x  is  less  than  *  /—  the  quantity  within  the  parentheses  of 

3  /  V 

equation  (21)  is  negative.     If  x  is  greater  than  -1/5—  this  quan- 

3   /  V 

tity  is  positive.     For  all  positive  values  of  x  less  than  \IJT-  the 

\  iir 

slope  is  negative,  and  the  curve  representing  y  is  descending;  for 

3   /   V 

values  of  x  greater  than  \f^-  the  curve  is  ascending.     Thus, 

\  2?r 

8/7 

the  surface  is  a  minimum  when  x  =  \l^-  •     From  equation  (16) 


=  \  —  , 

\   7T 


whence  --  =  4--     Since  the  v    does  not  appear  in 

Z 

this  equation,  the  ratio  of  x  to  z  is  constant,  ^,  for  all  values  of  v. 
The  most  economical  proportions  are  such  that  the  altitude  is 
equal  to  the  diameter  of  the  base. 

The  method  used  in  the  three  above  cases  is  applicable  to  any 
problem  where  the  functional  relation  between  the  two  variable 
quantities  x  and  y  is  given  by  means  of  an  equation. 

Exercises 

1.  Find  the  most  economical  dimensions  of  a  pan  with  square  base 
and  vertical  sides,  if  it  is  to  hold  5  gallons   (1  gallon  =  231   cubic 
inches). 

2.  Find  the  most  economical  proportions  for  a  cylindrical  can, 
open  at  the  top,  if  it  is  to  hold  1  quart  (1  quart  =  57f  cubic  inches). 

3.  What  is  the  ratio  of  the  length  of  the  legs  of  a  right  triangle 
when  its  area  is  a  maximum,  if  the  hypotenuse  is  a  constant,  10? 

4.  The  strength,  y,  of  a  rectangular  beam  varies  as  the  product  of 
its  breadth,  2,  by  the  square  of  its  depth,  x;  find  the  dimensions  of  the 
strongest  beam  which  can  be  cut  from  a  log  20  inches  in  diameter. 


CHAPTER  XIII 


[EMPIRICAL  EQUATIONS] 

143.  Empirical  Equation  Defined.  A  series  of  readings  con- 
necting two  or  more  variable  numbers  should  always  be  plotted 
upon  rectangular  coordinate  paper  (squared  paper),  thus  rep- 
resenting to  the  eye  the  variation  of  one  of  the  numbers  with 
respect  to  the  other. 

In  Table  VI,  there  are  given  seven  readings  of  the  velocity 
of  water  through  a  sample  of  sand.  These  seven  velocities 
correspond  each  to  a  measured  hydraulic  gradient.  This  set  of 
readings  when  plotted  upon  squared  paper  is  shown  in  Fig.  115. 


Velocity-  Feet  24  Hours 

i-.  to  co  *>.  01  o  .j 

OOOOOOOc 

C 

4 

+ 

% 

^ 

.'* 

^ 

.," 

'"^ 

X 

^ 

X1 

' 

x* 

• 

3 

£X 

•' 

*r' 

-' 

^ 

r+ 

X- 

*'  L 

10        20        30        40        50        60        70        80        90        100      110     120 
Hydraulic  Gradient-Feet  per  1000  Feet 

FIG.  115. 

The  plotted  points  do  not  lie  upon  a  smooth  curve.  The  reason 
for  this  is  that  there  are  errors  of  observation — errors  both  in 
the  determinations  of  the  velocities  and  in  the  hydraulic  gradients. 
If  there  had  been  no  errors  of  observation  and  if  all  pairs  of  read- 
ings h'ad  been  taken  under  exactly  the  same  conditions,  the  plotted 

264 


§143] 


MAXIMA  AND  MINIMA 


265 


points  would  have  fallen  upon  a  smooth  curve.  This  smooth 
curve  which  we  are  trying  to  locate  by  the  plotted  points  represents 
graphically  the  law  of  nature  which  we  are  trying  to  determine 
or  study,  viz.,  the  law  governing  the  flow  of  water  through  this 
particular  sample  of  sand,  relative  to  the  hydraulic  gradient. 

From  Fig.  115  it  is  apparent  that  the  plotted  points  do  not 
justify  the  assumption  of  a  locus  other  than  a  straight  line  along 
which  the  points  would  fall  if  there  were  no  errors  of  observation. 
The  true  relation  connecting  the  variable  numbers  measured 
may  not  be  the  linear  law,  but  with  the  data  at  hand  this  is  the 
only  law  which  can  be  assumed  as  an  approximation  to  the  law 
of  nature.  With  the  aid  of  a  transparent  triangle  the  line  QA 


Velocity-  Feet  per  Second 

to  CO  CO  CO  CO 
05  0  J-  JO  OS 

~-~  —  ' 

•r1 

> 

X 

f* 

xt 

I 

) 

^ 

\ 

> 

I 

\ 

\ 

\ 

1 

\ 

\ 
\ 

\ 

.1       .2        .3        .4        .5       .6         7 
Depth  -Part  of  Total 

FIG.  116. 


.8       .9 


is  drawn  among  the  plotted  points.  The  position  of  the  line  is 
located  by  the  eye  as  the  probable  line  upon  which  the  points 
would  fall  if  the  true  law  were  linear  and  if  there  were  no  errors  of 
observation.  The  equation  of  this  line  is  then  y  =  ax  -f-  b, 
where  x  represents  the  hydraulic  gradient  and  y  the  velocity. 
Since  the  line  passes  through  the  origin,  b  is  zero,  a  is  found  to 
be  0.51.  The  equation  of  the  line  becomes  y  =  0.51x,  and  is 
called  an  empirical  equation  connecting  velocity  and  hydraulic 
gradient. 


266 


MATHEMATICS 


[§143 


Frequently  it  is  desirable  to  express  the  relation  existing  be- 
tween variable  numbers  in  the  form  of  an  equation,  even  though 
this  equation  represents  only  an  approximation  to  the  true 
law.  Such  an  approximate  equation  when  found  by  means  of 
experimental  readings  is  called  an  empirical  equation. 

As  a  second  illustration  of  a  method  of  finding  the  empirical 
equation,  consider  the  plotted  points  in  Fig.  116.  From  this 
figure  it  is  seen  that  the  law  connecting  velocity  and  depth  is  not 
linear,  but  appears  to  be  parabolic.  The  vertex  of  the  parabola 
is  at,  or  nearly  at,  the  point  (0.29,  3.262)1.  If  the  origin  be  moved 
to  this  point  the  readings  (which  are  given  in  Table  IX)  become: 


X' 

y      \\      x1 

y 

-  0.29 

-  0.067 

0.21 

-  0.034 

-  0.19 

-  0.032 

0.31 

-  0  .  082 

-  0.09 

-  0.009 

0.41 

-  0.135 

0.01 

-  0.001 

0.51 

-  0.203 

0.11 

-  0.010 

0.61 

-  0.286 

and  the  equation  connecting  x'  and  y'  is  y'  =  a(x')2. 
If  (x')2  be  replaced  by  X  the  equation  becomes 

y'  =  aX  (1) 

Hence  if  y"s  are  plotted  against  X's,  the  squares  of  the  x"s, 
a  straight  line  will  be  obtained  if  the  curve  in  Fig.  116  is  a  parabola 
with  its  vertex  at  the  point  (0.29,  3.262).  These  plotted  points 
are  represented  in  Fig.  117.  The  points  indicated  by  circles  cor- 
respond to  [positive  values  of  x',  while  those  indicated  by  dots 
correspond  to  negative  values.  A  straight  line  drawn  among  the 
points  with  the  aid  of  a  transparent  triangle  gives  the  equation 

(2) 
(3) 


Since  X=  (x'}' 


Since 


and 


y'  =  -  0.77Z. 
y'  =  -  0.77(z')2. 


x'  =  x  -  0.29 


y'  =  y  -  3.262, 


1From  Fig.  116  it  may  at  first  appear  that  the  vertex  of  the  parabola  is  on  the 
line  x  =  0.3;  but  by  noting  the  vertical  positions  of  the  points  plotted  equidistant 
from  this  line  it  is  seen  that  the  line  of  symmetry  of  the  curve  is  slightly  to  its  left. 


§143] 


EMPIRICAL  EQUATIONS 


267 


equation  (3)  becomes 

y  -  3.262  =  -  0.77(a;  -  0.29)2, 
or 

y  =  -  0.77(z  -  0.29)2  +  3.262, 
or 

y  =  -  0.77  xz  +  0.447  x  +  3.197, 

an  empirical  equation  connecting  velocity  with  depth.1 


(4) 
(5) 


y 

-0.3 

/° 

/ 

, 

/ 

' 

.A 

V 

/ 

/ 

/ 

/ 

< 

to 

0 

/ 

0.64      0.08      0.12      0.16      0.20      0.24      0.28      O.S2       0.36     X 

FIG.  117. 

In  the  first  illustration  above  the  empirical  equation  is  of 
the  form  y  =  ax  +  b,  in  the  second  illustration  it  is  of  the  form 

1  Compare  equation  (4)  with  the  equation  given  in  exercise  4,  §  31. 


268  MATHEMATICS  [§144 

y  =  ax2  +  bx  +  c.  These  equations  are  each  a  special  case  of  the 
more  general  equation 

y  =  a0x»  +  aiZ"-1^-  azxn-z+  .   .    .  +  an_iz  +  an,         (6) 

where  a0, 01, 02,  .  .  . ,  an  are  coefficients  to  be  determined  and  where 
n  is  a  positive  integer.  To  determine  the  n  +  1  coefficients  of 
equation  (6),  n  +  1  equations  are  required.  Hence  the  curve 
corresponding  to  (6)  may  be  made  to  pass  through  (n  +  1) 
points  of  the  curve  drawn  among  the  plotted  points.  Empirical 
equations  of  the  form  (6)  are  undesirable  if  more  than  three  or 
four  terms  are  required. 

Exercises 

1.  From  the  data  given   in   Table   VII,  find  empirical  equations 
connecting  yield  and  head,  for  the  8-inch  and  14-inch  wells. 

2.  From  the  data  given  in  Table  VIII,  find  an  empirical  equation 
connecting  T  and  W. 

144.  Coefficients  Determined  by  the  Method  of  Least  Squares.1 
In  Fig.  115  a  linear  law  was  assumed  connecting  velocity  and 
hydraulic  gradient.  A  straight  line  was  drawn  among  the  points 
and  the  values  of  the  coefficients  a  and  b  were  determined  from  this 
line.  If  the  plotted  points  had  been  so  scattered  that  it  would 
have  been  difficult  to  estimate  by  eye  the  position  of  the  line,  the 
coefficients  a  and  6  could  have  been  determined  by  the  method  of 
least  squares.  This  method  will  be  illustrated  by  using  the  two 
illustrations  of  the  preceding  section.  In  the  first  illustration  the 
equation  is  assumed  to  be  y  —  ax  —  b  =  0.  Upon  substituting 
the  seven  pairs  of  values  given  in  Table  VI  there  results: 

16.9  -  31.9o  -6  =  0 

11.4  -  20. 8a  -6  =  0 

24.5  -  54.  la  -  6  =  0 
4.3-  10.0a-6  =  0 

10.1  -  21. Oa  -6  =  0 
58.8  -  119. la  -6  =  0 
22.7  -  55. 8a  -6  =  0 

1  The  student  will  verify  the  numerical  work  of  this  section. 


§144]  EMPIRICAL  EQUATIONS  269 

Here  are  seven  equations  from  which  to  solve  for  a  and  &.  This 
can,  of  course,  not  be  done,  since  the  number  of  equations  is  greater 
than  the  number  of  unknowns.  The  values  of  a  and  b,  which 
are  regarded  as  the  best  approximations,  are  found  as  follows: 
First  multiply  each  member  of  each  equation  by  the  coefficient  of 
a  in  that  equation  and  add  the  resulting  seven  equations.  In 
the  illustration  this  gives 

10627.52  -  22216.51a  -  312.76  =  0.  (1) 

Next  multiply  each  member  of  each  equation  by  the  coefficient  of 
6  in  that  equation  and  add.  In  the  illustration  this  gives 

148.7  -  312.7a  -  76  =  0.  (2) 

Next  solve  equations  (1)  and  (2)  for  a  and  b.  This  gives  in  the 
illustration 

a  =  0.48 
and 

6  =  -  0.2 

Thus,  the  equation  of  the  straight  line  drawn  among  the  points 
determined  by  this  method  is 

y  =  0.48z  -  0.2 
The  equation  found  by  the  method  of  the  preceding  section  is 

y  =  O.Slz 

In  the  second  illustration  (See  Fig.  116)  the  empirical  equation 
was  assumed  to  be  of  the  form  y  —  ax1  —  0£  —  7  =  0.  Upon 
substituting  the  ten  pairs  of  values  given  in  Table  IX  there 
results; 

3.195  -  0.00  a-  0.0/3  -7  =  0 

3.230  -  0.01  a  -  0.1/3  -  7  =  0 

3.253  -  0.04  a  -  0.2/3  -7  =  0 

3.261  -  0.09  a  -  0.30  -7  =  0 

3.252  -  0.16  a-  0.40  -7  =  0 

3.228  -  0.25  a  -  0.50  -  7  =  0 

3.181  -  0.36  a-  0.60  -7  =  0 

3.127  -  0.49  a-  0.7/3  -7  =  0 

3.059  -  0.64  a  -  0.80  -7  =  0 

2.976  -  0.81  a  -  0.90  -  7  =  0 


270  MATHEMATICS  [§145 

To  find  the  "best  values"  for  a,  /3  and  7  from  these  equations, 
first  multiply  each  member  of  each  equation  by  the  coefficient  of 
a  in  that  equation  and  add  the  resulting  ten  equations.  There 
results 

8.8289  -  1.5333a  -  2.025/3  -  2.867  =  0.  (3) 

Next  multiply  each  member  of  each  equation  by  the  coefficient  of 
/3  in  that  equation  and  add  the  resulting  equations.  There 
results 

14.09  -  2.025a  -  2.85/3  -  4.57  =  0.  (4) 

Next  multiply  each  member  of  each  equation  by  the  coefficient  of 
7  in  that  equation  and  add  the  resulting  ten  equations.  There 
results 

31.762  -  2.85a  -  4.5/3  -  107  =  0.  (5) 

Next  solve  equations  (3),  (4),  and  (5)  for  a,  /3,  and  7,  and  obtain 
a  =  -  0.81 
/3  =       0.48 
7=       3.19 

Hence  an  empirical  equation  found  by  this  method  is 
y  =  -  0.81x2  +  0.48z  +  3.19 

This  equation  places  the  vertex  of  the  parabola  at  the  point 
(0.296,  3.26). 

These  two  illustrations  show  that  a  large  amount  of  numerical 
work  is  required  to  find  the  coefficients  by  the  method  of  least 
squares.  If  the  plotted  points  fall  upon,  or  nearly  upon,  a  smooth 
curve  the  methods  of  the  preceding  section  should  be  used.1 

145.  Logarithmic  Paper.  The  student  will  read  §  73  again. 
In  Fig.  118  the  square  ADCB  is  one  unit  on  a  side.  At  the  bottom 
and  at  the  left-hand  side  of  this  square  are  logarithmic  scales,  each 
ranging  from  1  to  10.  Thus  on  the  lower  scale,  AD,  the  numbers 
are  placed  opposite  their  logarithms  measured  on  the  uniform 
scale  CB.  For  example,  1  is  placed  opposite  0;  2  opposite  0.301; 
3  opposite  0.477;  4  opposite  0.602;  5  opposite  0.699;  6  opposite 
0.778;  7  opposite  0.845;  8  opposite  0.903;  9  opposite  0.954, 
and  10  opposite  1.  The  distance  from  A  to  the  position  of  any 
number  on  the  scale  AD  is  equal  to  the  logarithm  of  that  number. 

1  The  theory  upon  which  the  method  of  least  squares  is  based  may  be  found 
in  texts  on  "Least  Squares." 


§145] 


271 


Through  the  points  of  division  of  the  AD  and  AC  scales,  vertical 
and  horizontal  lines  are  drawn  forming  what  is  called  logarithmic 
coordinate  paper.1  To  illustrate  the  use  of  logarithmic  coordinate 


0        .1 

1 

.2        .3 
I 

.< 

i 

.5 

i 

.£ 

: 

8 

.9 

LOg 

,  _  9 

X>    p 

^f 

s" 

^L 

x^ 

* 

** 

7 

^ 

^ 





-^ 

<L 

^ 

n 

t 

~2L 

-.5 

,/' 

4 

3 

-.2 

-J. 
.      0 

Al  2  MZ  4          56789  10  D 

FIG.  118. — A  single  square  (reduced)  of  logarithmic  paper. 

paper  for  determining  empirical  equations,  consider  the  data 
given  in  the  accompanying  table.  These  data  are  plotted  upon 
squared  paper  in  Fig.  119,  and  upon  logarithmic  paper  in  Fig.  118. 


X 

y 

X 

y 

1.5 

3.05 

6.5 

6.40 

2.5 

3.92 

7.5 

6.85 

3.5 

4.65 

8.5 

7.25 

4.5 

5.30 

9.5 

7.70 

5.5 

5.82 

'For  the  work  in  this  chapter  the  student  should  provide  himself  with  Single 
Logarithmic  cross-section  paper,  form  Af4,  Semi-logarithmic  coordinate  paper, 
form  M5,  and  Multiple  Logarithmic  coordinate  paper,  form  M 6. 


272 


MATHEMATICS 


[§145 


In  Fig.  118  the  plotted  points  lie  upon  a  straight  line  EF.  If  we 
imagine  the  logarithmic  rulings  to  be  removed  and  rulings  corres- 
ponding to  the  scales  CB  and  BD  to  be  drawn,  the  line  EF  would 
then  appear  upon  a  sheet  of  squared  paper  with  the  origin  at  A 
and  the  point  (1,  1)  at  B.  The  equation  of  the  line  EF  is  then 
y  =  aX  +  b,  where  X  and  Y  are  the  rectangular  coordinates  of  any 
point,  P,  on  the  line.1  a,  the  slope  of  the  line,  is  found  by  measure- 
Y 


^ 

^ 

^ 

^ 

^ 

X 

s* 

X 

/ 

S 

0123456789 

FIG.  119. 

ment,2  to  be  \.  b,  the  ^-intercept  AE,  is  found  from  the  scale  DB 
to  be  equal  to  0.398,  which,  from  the  scale  AC,  is  seen  to  be  equal 
to  log  2.5.  The  equation  of  the  straight  line  EF  is  then 

7  =  iZ  +  log2.5  (1) 

In  this  equation  Y  is  the  distance,  PM,  of  any  point,  P,  from  the 
line  AD;  but  this  distance  is  the  logarithm  of  y.     Similarly,  X,  the 
distance  PN,  is  the  logarithm  of  x.     Hence  equation  (1)  becomes 
log  y  =  £log  x  +  log  2.5, 

1X  and  Y  are  used  to  represent  the  rectangular  coordinates  of  a  point  in  contra- 
distinction to  the  x  and  y  when  the  logarithmic  scales  are  used. 

2  The  student  is  supposed  to  reproduce  Fig.  118  on  a  sheet  of  logarithmic  paper, 
form  Af4. 


§145] 
or 
or 
or 


EMPIRICAL  EQUATIONS 


273 


I 

10 
9 

S 
7 
G 
5 
4 
3 
2 
1 

log  y  =  log  x^-  +  log  2.5, 
log  y  =  log  (2.5x^2), 

( 

\ 

y 

/ 

/ 

/ 

. 

1 

/ 

(2) 


FIG.  120. 

an  empirical  equation  connecting  the  x  and  y  values  given  in  the 
above  table. 

18 


274 


MATHEMATICS 


[§145 


As  a  second  illustration,  consider  the  data  given  in  the  accom- 
panying table.  These  data  are  shown  plotted  upon  squared  paper 
in  Fig.  120,  and  upon  logarithmic  paper  in  Fig.  121.  Upon  the 


E 


X 

:  v     \ 

1          * 

y 

1.2 

2.15 

2.0 

5.90 

1.3 

2.50 

2.3 

7.80 

1.5 

3.35 

2.5 

9.30 

1.7 

4.30 

CO        .1 
10| u 


\ 


-.4 


-0 
A\  2  3  ^          5       6      1     8    9    10;; 

FIG.  121. 

logarithmic  paper  the  curve  is  a  straight  line.  Hence  its  equation 
is  Y  =  aX  +  b.  The  a  is  found  by  measurement  to  be  2.  The  b 
is  seen  to  be  log  1.5.  Then 


or 


Y  =  2X  +  log  1.5, 
;  y  =  2 logs  +  log  1.5, 


§145] 
or 
or 
or 


EMPIRICAL  EQUATIONS 

logy  =  logz2  +  log  1.5, 
log  y  =  log  1.5x2, 
y  =  1.5z2, 


275 


an  empirical  equation  connecting  the  x-  and  y-values  of  the  above 
table. 


As  a  third  illustration  consider  the  data  given  in  the  accom- 
panying table.  These  data  are  shown  plotted  upon  squared  paper 
in  Fig.  122  and  upon  logarithmic  paper  in  Fig.  121.  The  plotted 
points  in  Fig.  121  lie  upon  the  straight  line  GH,  whose  equation  is 
Y  =  aX  +  b.  a  is  found  to  be  —  1,  and  6  the  log  1.5.  Hence 

Y  =  -X+  log  1.5, 


276 


MATHEMATICS 


or 


or 


or 


or 


X 

.  V 

x                    y 

1.5 

10.0 

4.5 

3.30 

2.0 

7.5 

5.0 

2.98 

2.5 

6.0 

6.0 

2.49 

3.0 

5.0 

7.0 

2.12 

3.5 

4.25 

8.0 

1.87 

4.0 

3.73 

9.0 

1.65 

log  y  -  —  log  x  +  log  1.5, 


log  y  =  log  -  +  log  1.5, 
sc 


log  y  =  log  -f- ' 


1.5 

= 


an  empirical  equation  connecting  the  x  and  y  values  of  the  table. 

From  the  above  illustrations  it  is  seen  that  if  plotted  points 
upon  logarithmic  paper  determine  a  straight  line,  an  empirical 
equation  connecting  the  numbers  plotted  is  of  the  form 

y  =  Kz», 

where  n  is  the  slope  of  the  line  as  it  appears  on  the  logarithmic 
paper,  and  where  K  is  the  reading  on  the  logarithmic  scale  of  the 
point  of  intersection  of  the  line  with  the  vertical  line  passing 
through  the  point  (1,  1).  Conversely,  any  equation  of  the  form 
y  =  Kxn  plotted  upon  logarithmic  paper  gives  a  straight  line. 
For,  by  taking  the  logarithm  of  each  member  of  the  equation  there 
results 

log  y  =  n  log  x  +  log  K, 
which  is  of  the  form  Y  =  nX  +  b. 

From  the  above  illustrations  it  is  seen  that  the  uniform  scales  at 
the  top  and  right-hand  edge  of  the  logarithmic  paper  are  not  used 
in  the  actual  determination  of  the  constants  in  the  empirical 
equation.  These  scales  are  placed  there  for  instructional  pur- 
poses _only. 


§146] 


EMPIRICAL  EQUATIONS 


277 


Exercises 

1.  Through  the  lower  left-hand  corner  of  a  sheet  of  logarithmic 
paper,  form  M 4,  draw  lines  making  angles  of  30°,  45°,  and  60°  with  the 
horizontal.     Find  the  equations  connecting  the  x  and  y  for  each  of 
these  lines. 

2.  Through  the  upper  right-hand  corner  of  a  sheet  of  logarithmic 
paper,  form  M 4,  draw  lines  making  angles  of  30°,  45°,  and  60°  with  the 
horizontal.     Find  equations  connecting  the  x  and  y  for  each  of  these 
lines. 

3.  Through  the  upper  left-hand  corner  of  a  sheet  of  logarithmic 
paper,  form  M  4,  draw  lines  making  angles  of  30°,  45°,  and  60°  with  the 
vertical.     Find  equations  connecting  the  x  and  y  for  each  of  these 
lines. 

4.  Through  the  lower  right-hand  corner  of  a  sheet  of  logarithmic 
paper,  form  Jl/4,  draw  lines  making  angles  of  30°,  45°,  and  60°  with  the 
horizontal.     Find  equations  connecting  x  and  y  for  each  of  these  lines. 

5.  From  a  sheet  of  logarithmic  paper,  form  M 4,  find  to  three  deci- 
mal places  the  logarithm  of  (a)  2.1;  (6)  3.6;  (c)  4.8;  (d)  5.2;  (e)  7.21; 
(/)  9.3;  (?)  9.8. 

6.  Find  an  empirical  equation  connecting  x  and  y  of  the  accom- 
panying table. 


X 

y 

1         x 

y 

1.5 

3.55 

4.0 

1.59 

2.0 

2.80 

5.0 

1.22 

2.5 

2.34 

6.0 

1.14 

3.0 

2.02 

7.  Draw  lines  upon  a  sheet  of  logarithmic  paper,  form  Af4,  repre- 
senting 


(a)  y  =  x2. 

(6)  y=x*. 

(c)  y  =  2x*. 

(d)  y  =  2x^. 

(e)  y  =    -. 


(/)    y  =  x 


(g)  y  = 
(h)  y  = 
(t)  y  = 
(7)  y  =  9.2z~?*. 


146.  Multiple  Logarithmic  Paper.     In  the  preceding  section 
logarithmic  paper  was  used  (form  M4)  in  which  each  of  the  two 


278 


MATHEMATICS 


[§146 


logarithmic  scales  extended  from  1  to  10,  or  the  corresponding 
uniform  scales  extended  from  0  to  1.  It  is  apparent  that  if  num- 
bers falling  beyond  these  limits  were  to  be  plotted,  the  coordinate 
paper  would  have  to  be  extended.  Fig.  123  represents  a  sheet  of 


100 
90 
80 
70 
GO 
GO 

40 
30 

20 
10 

9 

8 
7 
6 
5 
4 

3 
2 

1 

A 

?                                               H                               Q             I 

f 

; 

/ 

/ 

I 

/ 

C 

•/ 

/ 

F, 

f 

f 

.  / 

/ 

/ 

/ 

1 

I 

/ 
/ 

/ 

/ 

D 

F 

1                2         3456  78910             20       30    40  C060        100 
K                  P                             M 

FIG.  123. — Four  squares  of  multiple  logarithmic  paper  (reduced). 

coordinate  paper  in  which  each  logarithmic  scale  extends  from  1  to 
100.     The  square  ADBC  is  the  same  as  the  square  ADBC  of  Fig. 
118,  or  of  form  M 4. 
Since 

log  20  =  log  10  +  log  2 
=  1  +  log  2 
=  AD  +  AK 
=  AM 
DM  =  AK 


Thus  the  vertical  line  marked  20  is  at  the  same  distance  from  D 
as  the  line  marked  2  is  from  A.     Similarly  the  vertical  lines 


§146] 


EMPIRICAL  EQUATIONS 


279 


marked  30,  40,  .  .  . ,  90,  and  100  are  at  the  same  distances  from 
D,  respectively,  as  the  lines  marked  3,  4,  .  .  .,9,  and  10  are 
from  the  point  A .  Similarly  the  portion  CG  of  the  vertical  scale  is 
the  same  as  the  portion  AC  excepting  that  the  numbers  attached 
to  the  upper  portion  are  ten  times  the  numbers  attached  to  the 
corresponding  points  on  the  lower  portion.  By  the  addition 
of  squares  like  ADBC  the  logarithmic  paper  may  be  extended 
over  any  range  of  values  on  the  horizontal  or  vertical  scales. 

It  is  apparent  from  an  examination  of  Fig.  123  that  the  point 
(1,  1)  (the  origin)  may  be  taken  at  any  corner  of  a  square.  Thus, 
if  B  is  taken  as  origin,  the  point  A  is  (0.1,  0.1)  and  /  the  point 
(10,  10);  if  H  is  taken  as  the  origin,  A  is  the  point  (0.1,  0.01) 
and  I  the  point  (10,  1).  In  this  work  it  is  supposed  that  the 
student  has  a  supply  of  multiple  logarithmic  coordinate  paper, 
form  M6. 

It  is  apparent  that  with  logarithmic  coordinate  paper  the  origin, 
i.e.,  the  point  (1,  1),  may  be  taken  only  at  the  corners  of  the 
squares,  while  with  squared  paper  the  origin,  i.e.,  the  point  (0,  0), 
may  be  taken  at  any  convenient  point. 

To  illustrate  the  use  of  multiple  logarithmic  paper  consider  the 
data  given  in  the  accompanying  table.  The  plotted  points  cor- 
responding to  these  numbers  are  shown  in  Fig.  123.  In  plotting 
these  points  the  point  B  is  taken  as  the  origin.  The  slope  of  the 
line  PQ  is  2  and  the  intercept  is  log  0.4.  Hence  the  equation  con- 
necting the  x-  and  ^/-values  of  the  table  is  y  =  0.4x2. 


X 

y       II       x 

y 

0.5 

0.10 

2.0 

1.64 

0.7 

0.20 

3.0 

3.70 

0.9 

0.33 

4.0 

6.50 

1.5 

0.92 

5.0 

10.00 

Exercises 

1.  Find  an  empirical  equation  between  x'  and  y',  using  the  numbers 
given  in  the  table  on  page  266. 

2.  Find  an  empirical  equation  connecting  W  and  T,  using  the  data 
given  in  Table  VIII. 

HINT:    First  move  the  zero  to  correspond  to  4°  centigrade.    Let  T' 


280 


MATHEMATICS 


[§147 


be  the  temperature  referred  to  this  new  zero.     Find  an  equation  be- 
tween W  and  Tr,  and  then  replace  T'  by  T  -  4. 

3.  Find  an  empirical  equation  connecting  I  and  W,  from  the  data 
given  in  Table  XI. 

147.  Translating  the  Curve  on  Logarithmic  Paper.  Consider 
the  data  given  in  the  accompanying  table.  These  data  plotted 
upon  multiple  logarithmic  paper  are  represented  by  small  circles 
in  Fig.  124.  The  plotted  points  do  not  lie  upon  a  straight 


5 

3 
2 

1 

0.5 

0.3 
0.2 

0.1 

\ 

V 

\ 

\ 

\ 

,  \ 

V 

\ 

\ 

\ 

\ 

*s 

\ 

.  V 

\ 

\ 

\ 

°\ 

\ 

\ 

\ 

( 

\ 

t 

\ 

i 

N 

• 

X 

0.1 


0.2      0.3 


0.5  1 

FIG.  124. 


5    E       10 


line,  but  if  they  are  translated  one  unit  to  the  right  they  fall  upon 
the  straight  line  EF.  Moving  the  points  one  unit  to  the  right  is 
equivalent  to  plotting  y  against  x  +  1.  Let  x  +  1  be  represented 
by  x'.  From  the  line  EF,  the  equation  connecting  y  and  x'  is 
found  to  be 

y  =  1.05(x')-*.  (1) 

Since  x'=  x  —  1,  equation  (1)  gives 

y  =  1.05(x  -  1)-*, 
an  empirical  equation  connecting  x  and  y. 


§148] 


EMPIRICAL  EQUATIONS 


281 


X 

y       i 

X 

y 

0.1 

8.00 

2.0 

0.40 

0.2 

4.75 

2.5 

0.31 

0.3 

3.30 

3.0 

0.25 

0.5 

1.95 

4.0 

0.17 

0.7 

1.3$ 

5.0 

0.13 

1.0 

0.90 

6.0 

0.105 

1.5 

0  57 

Frequently  it  may  happen  that  by  translating  the  curve  in  the 
x  direction  or  in  the  y  direction  the  plotted  points  upon  logarith- 
mic paper  will  lie  upon  a  straight  line. 

Exercise 

1.  By  plotting  upon  logarithmic  paper,  form  M6,  find  an  empirical 
equation  connecting  volume  and  pressure,  from  the  data  given  in 
Table  X. 

148.  Equations  Plotted  upon  Logarithmic  Paper.  If  a  large 
number  of  readings,  say  x,  are  to  be  substituted  in  a  formula  giving 
a  second  number,  say  y,  it  may  be  desirable  to  plot  a  curve  repre- 
senting the  equation  connecting  x  and  y,  and  from  this  curve  to 
scale  off  the  values  of  y  corresponding  to  the  readings  of  x.  If  this 
equation  plotted  upon  logarithmic  paper  gives  a  straight  line  it  is 
an  easy  matter  to  construct  the  curve.  Thus,  as  an  illustration, 
consider  the  formula 

Q  =  3.37Ltf^,  (1) 

giving  the  discharge  Q,  in  cubic  feet  per  second  (second-feet),  over 
a  trapezoidal  weir.  L  is  the  length,  in  feet,  of  the  crest  of  the  weir, 
and  H  is  the  depth,  in  feet,  of  the  water  on  the  crest.  This  formula 
for  various  values  of  L  will  give  on  logarithmic  paper  a  series  of 
parallel  lines  of  slope  f . 

Lines  for  L  equal  to  1,  2,.  3,  4,  and  5  are  given  in  Fig.  125. 

If  only  one  curve  of  the  family  is  to  be  drawn,  as  the  line  for 
L  =  2,  it  is  advisable  to  use  a  single  square  of  logarithmic  paper, 
form  M4.  From  Fig.  125  it  is  seen  that  the  line  L  =  2  crosses  six 
squares.  Instead  of  using  these  six  squares  a  single  square  may  be 
traversed  six  times  as  shown  in  Fig.  126.  The  segments  of  the  line 


282 


MATHEMATICS 


[§148 


in  Fig.  126  are  lettered  the  same  as  in  Fig.  125.     The  segment  AB 
is  not  shown  in  Fig.  126  as  it  coincides  with  the  segment  F2. 

The  terms  tenths,  units,  and  tens  written  above  the  lines  indicate 
the  denomination  of  the  figures  on  the  vertical  scale.     Similarly 


L 

5432     1 


10 


1.0 


o.oi 


lii\ 


// 


s 


M 


Z 


a 

o.i  i  10 

FIG.  125. — Curves  for  Q  =  3.37L.ff2£  plotted  upon  logarithmic  paper. 


the  terms  hundredths,  tenths,  etc.,  written  below  the  lines  indicate 
the  denomination  of  the  figures  on  the  horizontal  scale.  Thus,  to 
illustrate,  if  the  reading  of  H  is  0.7,  the  line  segment  DE  is  used, 
and  from  it  and  the  vertical  scale  the  discharge,  Q,  is  found  to  be 
3.95  second  feet. 


§149] 


EMPIRICAL  EQUATIONS 
Exercises 


283 


1.  Upon  a  sheet  of  logarithmic  paper,  form  M6,  draw  a  series  of 
curves  for 

Q  =  3. 
for  L  =  1,  2,  3,  4,  and  5. 


z 


7 


^ 


1     8    9  10 


FIG.  126.— Curve  for  Q=3.37-2HM  plotted  upon  one  square  of  log- 
arithmic paper. 


2.  Upon  a  sheet  of  logarithmic  paper,  form  M 4,  draw  a  series  of 
line  segments  for 

Q  =  3.37LH* 
if  L  =3,  and  if  H  ranges  from  0.1  to  10. 

149.  Semi-logarithmic  Paper.  Semi-logarithmic  paper,  form 
Mb,  is  coordinate  paper  ruled  logarithmically  in  one  direction  and 
uniformly  in  the  other. 


284 


MATHEMATICS 


[§149 


The  data  given  in  the  accompanying  table,  plotted  upon  semi- 
logarithmic  paper,  is  shown  in  Fig.  127. 


X 

y 

0.2 

3.18 

0.4 

3.96 

0.6 

5.00 

0.8 

6.30 

or 


0         0.1       0.2        0.3        0.4        0.5       0.6        0.7       0.8         0.9        1 

FIG.  127. — Semi-logarithmic  paper  (reduced). 

The  equation  of  this  straight  line  is 

,      Y  =  fcc  +  log  2.51, 

log  y  =  \x  +  log  2.51, 


or 


log 


M       x 
2.51      2 


§149] 
or 

or 


EMPIRICAL  EQUATIONS 


y 

2.51 


285 


y  =  2.51(10)z/2, 
an  empirical  equation  connecting  the  x-  and  ^-values  of  the  table. 

Exercises 

1.  Find  an  empirical  equation  connecting  the  x-  and  the  y-values 
given  in  the  accompanying  table. 


X 

y 

0.2 

5.8 

0.4 

4.4 

0.6 

3.4 

0.8 

2.6 

APPENDIX 
GREEK  ALPHABET 


LETTERS 

NAMES 

LETTERS    NAMES 

LETTERS 

NAMES 

A 

a 

Alpha 

I 

4 

Iota 

P 

P 

Rho 

B 

ft 

Beta 

K 

* 

Kappa 

2 

ff,    S 

Sigma 

r 

y 

Gamma 

A 

X 

Lambda 

T 

T 

Tau 

A 

d 

Delta 

M 

n 

Mu 

T 

V 

Upsilon 

E 

6 

Epsilon 

N 

t 

Nu 

4> 

(f),<p 

Phi 

Z 

r 

Zeta 

H 

{ 

Xi 

X 

X 

Chi 

H 

n 

Eta 

O 

o 

Omicron 

^ 

t 

Psi 

e 

6,& 

Theta 

n 

* 

Pi 

n 

0} 

Omega 

a  X  b,  or  a-b,  or  ab 
a  -T-  b,  or  a :  b,  or  -r 


a^ 
a3 
a4 
a"  =  a  a  a  .  .a 


a^  =  Va 
a1/*  =\/a 
a°  =  1 


SYMBOLS 

plus. 

minus. 

plus  or  minus. 

minus  or  plus. 

equals. 

a  times  b. 

a  divided  by  b. 

parentheses,  brackets,  braces, 
a  square. 
a  cube 

a  to  the  fourth  power, 
a  to  the  nth,  or,  a  nth  (n  a's  multi- 
plied together,  if  n  is  a  positive 
integer), 
square  root  of  a. 
cube  root  of  a. 
rth  root  of  a. 


286 


APPENDIX  287 

a',  a",  a'",   .    .    .,  a"  (accents)    read,  respectively,    "a 

prime,"  "a  second,"  "a  third," 
etc. 

Qo,  01,  a2,    .    .    . ,  a«  (subscripts)     read,    respectively, 

"a  sub  zero,"  "a  sub  one,"  "a 
sub  two,"  etc. 

7*  is  not  equal  to. 

>  is  greater  than. 

<  is  less  than. 

is  equal  to  or  greater  than, 
is  equal  to  or  less  than. 

.^  angle. 

\n_  (read,  "factorial  n")  means   the 

product  of  all  the  integral  num- 
bers from  1  up  to  and  including 
n.  Thus:  |5_  =  1-2-3-4-5. 

n\  (1_^  is  sometimes  written  n!). 

|a|  the  numerical,  or  absolute,  value 

of  a. 

/Or);  <t>(x);  F(x)',  ^(x),  etc.  function  of  x. 

logo  x  the  logarithm  of  x  to  the  base  a. 

log  x  the  logarithm  of  x  to  the  base  10- 

2  the  sum  of  such  terms  as 

i  —  ft  the  sum  of  such  terms  as,   .    .    ., 

2  asi  varies  from  a  to  /3. 

TT  the  ratio  of  the  length  of  the  cir- 

cumference of  a  circle  to  the 
length  of  its  diameter.  An  ap- 
proximate value  of  TT  is  22/7.  A 
better  approximation  is  3.1416. 

e  the  base  of  the  natural  (or  hyper- 

bolic)   logarithmic    system,    e  = 
2.71828  approximately, 
the  number  of  permutations  of 
n  things  taken  r  at  a  time. 

the  number  of  permutations  of  n 
things  taken  all  at  a  time. 


288 
»CV  = 

00 

lim 


_l!L 


In    In 


u.m    Y 

=     (* 


MATHEMATICS 


the  number  of  combinations  of  n 
things  taken  r  at  a  time. 

infinite. 

limit. 

approaches. 

the  limit  of  Y  as  x  approaches  a. 


FORMULAS 
Algebra 


5. 
6. 
7. 
8. 
9. 

10. 


~£~  a    =  d 


\am]n  =  a 
l"1 
J     = 


a)  (6)  =  -  (06). 
o)(-  6)  =  06. 
+  6)2  =  a2  +  2a6  +  b2. 
+  b)3  =  a3  +  3a26  +  3a62  +  b3. 
+  6)4  =  a4  ±  4a36  +  6a262  ±  4a63 

/  ••  \ 

±  - 


=  a"  + 


11.  a2  +  2a6  +  b2  =  (a  +  6)2. 

12.  a2  -  62  =  (a  +  6)  (a  -  6). 

13.  a3  -  b3  =  (a  -  6)(a2  +  ab  +  62). 

14.  a3  +b3  =  (a  +  b)  (a2  -  a&  +  62). 


64. 


n(n  -  l)(n  -  2) 

- 


15.  If 


c  =  0,  x  = 


16.  logo  (xy)  =  Iog0  x  +  logo  y. 

2C 

17.  logo  -  =  logo  x  -  logo  y. 

18.  Iog0  xn  =  n  logo  x. 

19.  loga  --    =    -  logo  X. 

£> 

20.  logo  a;  =  (logb  x)(loga  6). 


22.  log.  x  =  log€  10  logic  x  =  2.3026  logic  x. 

23.  logic  x  =  logic  e  log,,  x  =  0.4343  loge  x. 


APPENDIX 


289 


24.  a  +  [a  +  d]  +  [a  +  2d]  +  •    •    •    +  la  +  (n  -  l)d~] 


25.  a  +  ar  +  ar2  +  ar3  + 

26.  a  +  ar  +  ar2  +  ar3  + 

27.  1+2+3+4  +  •  • 

28.  I2  +  22  +  32  +  42  +  • 


+  ar""1  = 


to    oo     = 


•(»- 

-  r") 


1  -  r 


1  -  r 


>  if  |r|<  1. 


+  »  =  2  (n  + 


+  n2  =  a(n  +  l)(2n  +  1). 


29.  I3  +  23  +  33  +  4'  +  •   •  -    +  n3  =  fj-(n  +1)1* 

to  ~  =  Hm  r  i  +  1?  = 

»  *  «L         n 


so.  i  +  p  +  +  + 


31.  e*  =  1  +  x  +        +  ~ 


=  2.71828. 
to  oo    (for  all  values  of  x). 


32.  log,  (1  +  z)  =  +  x  -  -  +  -  +    ...  to  oo  (if  -  1<  x  <+  1). 

33.  e~x    =  1  —  x2  +  -jTj  —  ^  +    •    •    •  to  oo  (for  all  values  of  x). 

•   (n  -  r  +  1)  = 


34.  nPr  =  n(n  -  l)(n  -  2) 

35.  nPB  =  P»  =  n. 

36.  nPr  =  n(._iP,_i). 

37.  nCr    =    ~nPr      = 


n 

In  — *r 


Trigonometry 


38.  sin  a  esc  a  =  1. 

39.  cos  a  sec  a  =  1. 

40.  tan  a  cot  a  =  1. 

41.  sin2  a  +  cos2  a  =  1. 

42.  sec2  a  =  1  +  tan2  a. 

43.  esc2  a  =  1  +  cot2  a. 


,  . 
44. 


cos  a 
19 


=  tan  a. 


290  MATHEMATICS 

.     COS  a 

45.  —  -  =  cot  a. 
sin  a 

46.  sin  (90°  ±  a)  =  cos  a. 

47.  cos  (90°  +  a)  =  +  sin  a. 

48.  tan  (90°  ±  a)  =  +_cot  a. 

49.  sin  (180°  ±  a)  =  +  sin  a. 

50.  cos  (180°  ±  a)  =  -  cos  a. 

51.  tan  (180°  +  «)  =  ±  tan  a. 

52.  sin  (270°  ±  a)  =  -  cos  a. 
63.  cos  (270°  ±  a)  =  ±  sin  a. 
54.  tan  (270°  ±  o)  =  +  cot  a. 

56.  sin  (a  +  )3)  =  sin  a  cos  j3  +  cos  a  sin  /3. 
56.  cos  (a  ±  /3)  =  cos  a  cos  /3  +  sin  a  sin  /3. 

tan  a  ±  tan  (3 
67.  tan  («**)- 


58.  sin  2a  =  2  sin  a  cos  a. 

59.  cos  2a  =  cos2  a  —  sin2  a. 

60.  =1—2  sin2  a. 

61.  =  2  cos2  a  -  1. 

2  tan  or 


62.  tan  2a  = 


z  —    —  r 
1  —  tan2  a 


DO.  sin     £ 
64.  cos  K 

RK      tor,      a 

1      \ 

/I  +  COS  a 

'    \          2 

VI  —  COS  a 

bo.  tan  o 
66. 

CT 

1  +  COS  a 
1  —  COS  a 

sin  a 
sin  a 

a  +  /3  a  — 

68.  sin  a  +  sin  /?  =  2  sin  — s —  cos  — s- 

^  ^ 

a  +  /3          a  — 

69.  sin  a  —  sin  /3  =  2  cos  — o —  sm  " 

a  +  0 

70.  cos  a  +  cos  /3  =  2  cos  — o —  cos 

71.  cos  a  —  cos  /3  =  —  2  sin  — ^ —  sin  — ^ — • 

&  & 

72.  — —  =  — — a  =  ~ '  Law  of  sines. 

sm  or     sm  /3       sm  7 


APPENDIX  291 

73.  a2  =  62  +  c2  —  26c  cos  a.        Law  of  cosines. 

a-  j8 

tan       2  a  —  b 

74.  -; — ?•  Law  of  tangents. 
tan^3  a  +  b 

75.  Area  of  triangle  =  A/S(S  —  a)(s  —  6)(s  —  c), 

where  2s  =  a  +  6  +  c. 

cfe  sin  a 
7b.  =        2 

c2  sin  a  sin  8 
77. 


78.  Radius  of  inscribed  circle  =  A/ 

79.  sin2  a  =  |(1  —  cos  2a). 

80.  cos2  a  =  5(1  +  cos  2a). 

1  —  cos  2a 


1  +  cos  2a 
(1  -  cos  2a)2 
sin2  2  a 
sin2  2  a 


81.  tan2  a  = 
82. 

00  

(1  +  cos  2a)2 

84.  sin  3a  =  3  sin  a  —  4  sin3  a. 

85.  cos  3a  =  4  cos3  a  —  3  cos  a. 

3  tan  a  —  tan3  a 

86.  tan  6 a  =  — ^ o'+TTi — 

1  —  o  tan2  a 


a° 

87.  a  >  sin  a  >  a  —  -„-  if  a  is  expressed  in  radians. 

a3  a3          a& 

88.  a  —  -^  <  sin  a  <  a  —  -vr  H — ^    if  a  is  expressed  in  radians. 

o  o  O 

If  a  is  small  and  expressed  in  radians: 

89.  sin  a  =  a  approximately,  or 

a3 

90.  sin  a  =  a  —  r  _      is  a  better  approximation,  or 

o 


91.  sin  a  =  a  —    „•-   +  ~=-     is  a  still  better  approximation. 

a2 

92.  1  >  cos  a  >  1  —  ~~-      if  a  is  expressed  in  radians. 

a2  a2          a4 

93.  1  —  ;-„•  <  cos  a  <  1  —  rfc-  +  —r-   if  a  is  expressed  in  radians. 

I «  Le  4t 


292  MATHEMATICS 


a*    * 
94.  1  — 


*r 


if  a  is  expressed 
in  radians. 


If  a  is  small  and  measured  in  radians: 
95.  cos  a  =  1    approximately,  or 


96.  cos  a  =  1  —  TTT   is  a  better  approximation,  or 

a2         a4 

97.  cos  a  =  1  —  pr-  +  nr    is  a  still  better  approximation. 

\—     L*. 

Analytical  Geometry 

98.  y  =  az  +  b,  slope  equation  of  straight  line. 

99.  x  cos  a  +  y  sin  a  =  p,  normal  equation  of  straight  line. 

100.  x2  +  y2  =  r2,  equation  of  circle,  center  at  origin,  radius  r. 

101.  (x  —  a)2  +  (y  —  /3)2  =  r2,  equation  of  circle,  center  at  the  point 

(a,  /3),  radius  r. 

102.  y  =  px2,  equation  of  parabola  with  axis  coinciding  with  F-axis. 

103.  y  =  p(x  —  a)2,  equation  of  parabola  with  axis  parallel  to  the 

F-axis,  but  a  units  to  its  right  (to  its  left  if  a   is 
negative). 

x2       v2 

104.  — j  +  rj-  =  1,  equation  of  ellipse,  axes  coinciding  with  the  co- 

ordinate axes,  foci  upon  X-axis. 

105.  Area  of  the  ellipse  =  irab. 

2^2  4*2 

106.  — j  • — j-j  =  1,  equation  of  hyperbola,  axes  coinciding  with  co- 

ordinate axes. 

107.  x2  —  y-  =  a2,  equation  of  equilateral  hyperbola,  axes  coinciding 

with  coordinate  axes. 

108.  xy  =  k,  equation  of  equilateral  hyperbola  referred  to  asymptotes 

as  axes. 

109.  Ax  +  By  +  Cz  =  D,  equation  of  plane. 

110.  x  cos  a  +  y  cos  |8  +  z  cos  y  =  p,  normal  equation  of  plane. 

111.  x2  +  y2  +  z2  =  r2,  equation  of  sphere,  center  at  origin,  radius  r. 

112.  (x  —  a)2  +  (y  —  b)2  +  (z  —  c)2  =  r2,  equation    of  sphere,  cen- 

ter at  the  point  (a,  b,  c). 

113.  cos2  a  +  cos2  ft  +  cos27  =  1,  where  a,  0,  and  7  are  the  direction 

angles  of  a  line. 


APPENDIX 


293 


114.  cos  0  —  cos  «i  cos  at  +  cos  0i  cos  02  +  cos  71  cos  72,  where  0 
is  the  angle  between  two  lines  whose  direction  angles  are, 
respectively,  a\,  0i,  71,  and  «2,  02,  72. 

MENSURATION 


General  Triangle 

116.  Area  =  %bp. 

116.  a  +  0  +  7  =  180°. 

117.  5   =  a  +  0. 


^T 


Right  Triangle 


118.  62  +  a2  =  c2. 

119.  p2  =  mn. 

120.  b2  =  en. 


Circle 

r  =  radius. 

121.  Circumference  =  2-irr. 

122.  Area  =  Trr2. 

s  =  length  of  arc  ab. 

123.  Area  of  sector  =  %rs. 


Sphere 

radius  =  r. 
124.  Volume  =  f*-r3. 
126.  Area  of  surface  =  4irr2. 

h  =  altitude  of  zone. 

126.  Area  of  zone  abed  =  2*rh. 
a,    b   =  radii     of   bases    of 

segment. 

127.  Volume  of  segment  = 


294 


MATHEMATICS 


A.. 


Right  Circular  Cone 

r    =  radius  of  base. 

h   =  alitude. 

s    =  slant  height. 


s    =       r2  +  h2. 

128.  Volume  =  \irr*h. 

129.  Area     of     convex 

=  wrs. 


surface 


Frustum  of  Right  Circular  Cone 

R  =  radius  of  lower  base. 

r    =  radius  of  upper  base 

h   =  altitude. 

s    =  slant  height. 

s    =    V/i2  +  (R  -  r)\ 

130.  Area  of  convex  surface  = 
irs(R  -  r). 

131.  Volume 

=  ~(R^  +  Rr  +  r*). 


132. 


3(R  -  r) 


Anchor  Ring 
of 


generating 


r    =  radius 
circle. 
R  =  mean  radius. 

133.  Volume  =  2w*r*R. 

134.  Area  of  surface  =  4v*rR. 


APPENDIX 
Theory  of  Probability 


295 


h 
"" 


136.  y  =  — —e"h  *  ,  probability  curve. 
136.  <r   =  \/— ,  standard  deviation. 


137.  E  = 


n 
0.6745 


_ 
\/2/t>2,  probable  error  of  mean. 


TABLE  XVI.— DECIMAL  EQUIVALENTS  OF  COMMON  FRACTIONS 


8ths     |     16ths 

8ths    |     16ths 

1/8  =0.125 
1/4  =0.250 
3/8  =  0.375 
1/2  =0.500 

1/16  =  0.0625 
3/16  =  0.1875 
5/16  =  0.3125 
7/16  =  0.4375 

5/8  =0.625 
3/4  =0.750 
7/8  =0.875 

9/16  =  0.5625 
11/16  =  0.6875 
13/16  =  0.8125 
15/16  =  0.9375 

TABLE   XVII.— CONVERTING   INCHES  INTO   FEET 


Inches 

0 

1/8 

1/4     |     3/8 

1/2     |     5/8     |     3/4 

7/8 

0 

0.000 

0.010 

0.021 

0.031 

0.042 

0.052 

0.062 

0.073 

1 

0.083 

0.094 

0.104 

0.115 

0.125 

0.135 

0.146 

0.156 

2 

0.167 

0.177 

0.188 

0.198 

0.208 

0.219 

0.229 

0.239 

3 

0.250 

0.260 

0.271 

0.281 

0.292 

0.302 

0.312 

0.323 

4 

0.333 

0.344 

0.354 

0.365 

0.375 

0.385 

0.396 

0.406 

5 

0.417 

0.427 

0.437 

0.448 

0.458 

0.469 

0.479 

0.490 

6 

0.500 

0.510 

0.521 

0.531 

0.542 

0.552 

0.562 

0.573 

7 

0.583 

0.594 

0.604 

0.615 

0.625 

0.635 

0.646 

0.656 

8 

0.667 

0.677 

0.688 

0.698 

0.708 

0.719 

0.729 

0.740 

9 

0.750 

0.760 

0.771 

0.781 

0.792 

V.  802 

0.812 

0.823 

10 

0.833 

0.844 

0.854 

0.865 

0.875 

0.885 

0.896 

0.906 

11 

0.917 

0.927 

0.937 

0.948 

0.958 

0.969 

0.979 

0.990 

296 


MATHEMATICS 


TABLE   XVIII— BOAED   MEASURE 

Table  giving  the  number  of  board  feet  in  timbers   of  different 
dimensions;  1  board  foot  =  144  cubic  inches. 


Size  in  inches 

Length  in  feet 

8 

10      |      12      |      14 

16             18 

20 

1  X    4 

2f 

31 

4 

4! 

5i 

6 

61 

1  X    6 

4 

5 

6 

7 

8 

9 

10 

1  X    8 

5| 

61 

8 

9* 

lOf 

12 

13| 

1  X  10 

6f 

8| 

10 

ill 

13| 

15 

16§ 

1  X  12 

8 

10  ; 

12 

14 

16 

18 

20 

2X4 

51 

6f 

8 

»* 

10| 

12 

13* 

2X6 

8 

10 

12 

14 

16 

18 

20 

2X8 

10f 

13* 

16 

18f 

21| 

24 

26f 

2  X  10 

is* 

18f 

20 

23i 

26| 

30 

33£ 

2  X  12 

16 

20 

24 

28 

32 

36 

40 

3X4 

8 

10 

12 

14 

16 

18 

20 

3X6 

12 

15 

18 

21 

24 

27 

30 

3X8 

16 

20 

24 

28 

32 

36 

40 

3  X  10 

20 

25 

30 

35 

40 

45 

50 

3  X  12 

24 

30 

36 

42 

48 

54 

60 

4X4 

10f 

13* 

16 

18f 

21| 

24 

26f 

4X6 

16 

20 

24 

28 

32 

36 

40 

4X8 

21* 

26| 

32 

37| 

42| 

48 

53i 

4  X  10 

26f 

33| 

40 

46| 

53| 

60 

66f 

4  X  12 

32 

40 

48 

56 

64 

72 

80 

6X6 

24 

30 

36 

42 

48 

54 

60 

6X8 

32 

40 

48 

56 

64 

72 

80 

6  X  10 

40 

50 

60 

70 

80 

90 

100 

6  X  12 

48 

60 

72 

84 

96 

108 

120 

8X8 

42f 

53  * 

64 

74f 

85| 

96 

106f 

8  X  10 

53£ 

66| 

80 

93§ 

106f 

120 

133^ 

8  X  12 

64 

80 

96 

112 

128 

144 

160 

10  X  10 

66f 

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100 

116f 

133| 

150 

166! 

10  X  12 

80 

100 

120 

140 

160 

180 

200 

12  X  12 

96 

120 

144 

168 

192 

216 

240 

APPENDIX 


297 


TABLE  XIX— WATER  CONTAINED  IN   1   FOOT  LENGTH  OF  CIR- 
CULAR  PIPE   OF  d   INCHES    INSIDE   DIAMETER 


d 

Gallons 

Weight  Iba.    |              d 

Gallons 

Weight  Ibs. 

H 

0.010 

0.085 

7 

1.999 

16.683 

1                0.041 

0.340 

8 

2.611 

21.790 

1M             0.092 

0.766 

10 

4.080 

34  .  048 

2                0.163 

1.362 

12 

5.875 

49  .  028 

2H 

0.255 

2.128 

14 

7.997 

66.733 

3 

0.367 

3.064 

16 

10.44 

87.162 

SM 

0.500 

4.171 

18 

13.22 

110.314 

4 

0.653 

5.448   . 

20 

16.32 

136.190 

5                1  .  020 

8.512 

22 

19.75 

164.790 

6 

1.469 

12.257 

24 

23.50 

196.114 

TABLE    XX— TABLE  OF  HEADS  OF  WATER  CORRESPONDING  TO 

GIVEN  PRESSURE  AND  OF  PRESSURE  CORRESPONDING  TO 

GIVEN  HEAD 


Pressure,  Ibs.  per  sq.  in.|      Head,  feet      ||     Head,  feet 

Pressure,  Ibs.  per  sq.  in. 

1 

2.307 

1 

0.434 

2 

4.614 

2 

0.867 

3 

6.920 

3 

1.301 

4 

9.227 

4 

1.734 

5 

11.534 

5 

2.168 

6 

13.841 

6 

2.601 

7 

16.147 

7 

3.035 

8 

18.454 

8 

3.468 

9 

20.761 

9 

3.902 

10 

23.068 

10 

4.335 

20 

46.135 

20 

8.670 

30 

69.203 

30 

13.005 

40 

92.271 

40 

17.340 

50 

115.338 

50 

21.675 

60 

138.406 

60 

26.010 

70 

161.474 

70 

30.346 

80 

184.541 

80 

34.681 

90 

207.609 

90 

39.016 

100 

230.677 

100 

43.351 

298 


MATHEMATICS 


TABLE 

XXI.—  CONVERSION  TABLES 

Lengths 

1  centimeter 

=  0.3937 

inch. 

1  meter 

=  39.37 

inches. 

1  kilometer 

=  0.62137 

mile. 

1  inch 

=  2.540 

centimeters. 

1  foot 

=  30.480 

centimeters. 

1  yard 
1  mile  (5280  feet) 

=  0.9144 
=  1.609 

meter, 
kilometers. 

Areas 

1  square  centimeter 
1  square  meter 
1  square  kilometer 
1  square  inch 
1  square  foot 
1  square  mile 

=  0.1550 
=  10.764 
=  0.3861 
=  6.4516 
=  0.0929 
=  2.590 

square  inch, 
square  feet, 
square  mile, 
square  centimeters, 
square  meter, 
square  kilometers. 

Volume 

1  cubic  centimeter 

=  0.0610 

cubic  inch. 

1  cubic  meter 
1  liter 

1  cubic  inch 

=  1.308 
=  61.023 

=  16.387 

cubic  yards, 
cubic  inches  =  1  .  057  quarts 
(liquid), 
cubic  centimeters. 

1  cubic  foot 

=  7.480 

gallons  (liquid)  =  28.317 
liters. 

1  cubic  yard 
1  pint  (liquid) 
1  gallon 

=  0.7646 
=  28.875 
=  231 

cubic  meter, 
cubic  inches  =  0.4732  liter, 
cubic  inches  =  3.  785  liters. 

Weights 

1  gram 

1  kilogram 
1  grain 

=  15.432 

=  2.2046 
=  0.0022857 

grains     =     0.03527     ounce 
(Av.). 
pounds  (Av.). 
ounce    (Av.)     =     0.064799 

1  ounce  (Av.) 
1  pound  (Av.) 

=  437.5 
=  0.45359 

gram, 
grains  =  28  .  3495  grams, 
kilogram   =  7000  grains. 

APPENDIX 


299 


Pressure 

1  foot  of  water  column  =  0.4335 
=  22.419 
=  0.8826 
=  0.0295 

1  inch  of  mercury  column  =  1 . 133 
=  0.4912 
=  0.03342 

1  pound  per  square  inch 


1  atmosphere 


=  51.712 
=  2.307 
=  2.036 
=  760 
=  33,901 
=  29.921 
=  14.697 


pounds  per  square  inch. 

millimeters  of  mercury. 

inch  of  mercury. 

atmosphere. 

feet  of  water. 

pound  per  square  inch. 

atmosphere. 

millimeters  of  mercury. 

feet  of  water. 

inches  of  mercury. 

millimeters  of  mercury. 

feet  of  water. 

inches  of  mercury. 

pounds  per  square  inch. 


Weight  and  Volume  of  Water 


1  cubic  centimeter  water 
1  pint  (liquid)  water 
1  cubic  foot  water 
1  ounce  (Av.) 


=  1  gram. 

=  1 . 043  pounds. 

=  62.428  pounds  (Av.). 

=  28.35  cubic  centimeters  water. 

=  1.730  cubic  inches  water. 


1  cubic  foot  per  second 
1  acre  foot  per  day 


Flow  of  Water 
=  450 


=  0.5042 


gallons  per  minute  (approxi- 
mately), 
cubic  foot  per  second.  • 


1  part  per  million 
1  grain  per  gallon 


Salts  in  Solution 

=  0.05842      grain  per  gallon. 
=  17.118       parts  per  million. 


300 


MATHEMATICS 


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APPENDIX 


301 


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INDEX 


Abscissa  of  point,  33 

axis  of,  25 

Acceleration  of  falling  body,  51 
Accuracy    of    logarithmic    com- 
putation, 143 
Addition,  12 

formulas,  108 

vector,  156 

Aggregation,  symbols  of,  16 
Algebra,  formulas  of,  288 

review  of,  12 
Alphabet,  Greek,  286 
Analytic  geometry,  formulas  of, 

292 
Angle  of  depression,  86 

direction,  243 

of  elevation,  86 

initial  side  of,  80 

to  measure  an  arc,  3 

sense  of,  80 

terminal  side  of,  80 

unit  of,  91 

vertex  of,  80 
Angular  magnitude,  80 
Anti-logarithmic  tables,  73 
Anti-logarithms  found  from  loga- 
rithmic tables,  78 
Approximation  formulas,  196 
Area  of  ellipse,  136 

of  triangles,  71,  110,  111 
Arrangement       of      logarithmic 

work,  71,  74 
Axes,  coordinate,  25 

marking,  30 

of  the  ellipse,  133 


Axis,  X-,  33 
Y-,  33 

of  abscissas,  25 
of  ordinates,  25 
of  parabola,  25 

Bacteria,  growth  of,  .66 

Base  of  logarithmic  system,  61 

change  of  logarithmic,  66 
Binomial  coefficients,  195 

expansion,  189,  195 

for  negative  and  fractional 
exponents,  195 

theorem,  194,  195 
Biological  measurements,  229 
Board,  drawing,  2 
Boyle's  law,  58 

Charts,  precipitation,  30 
Characteristic  of  logarithm,  64 
Circle,  equation  of,  37 

orthographic   projection   of, 

138 

Circular  measure,  91 
Coefficients,  binomial,  195 

determined  by  the  method 

of  least  squares,  268 
Combinations,  189,  190,  193 

formula  for  number  of,  193 
Compasses,  3 

Compound  interest  law,  66,  197 
Compression,  179 

of  air,  58 

Computation,  accuracy  of  loga- 
rithmic, 143 


303 


304 


INDEX 


Concrete,  cost  of,  47 

mixture,  47 
Cone,  section  of.  139 
Coordinate  axes,  25 

paper,  25 

planes,  239 

x-,  34 

V,  34 

Coordinates  of  a  point,  33,  239 
Corrections  to  logarithms,  70 
Cosines,  direction,  243 

law  of,  105 
Cost  of  concrete,  47 
Couple,  170 

moment  of,  170,  171 
Count,  least,  231 
Crane,  simple,  164 
Cube  roots,  found  with  the  slide 

rule,  153 
Cross  section,  normal,  of  cylinder, 

115 

Current  meter,  54 
Curve,  decreasing,  52 

error,  224 

equation  of,  37 

increasing,  52 

of    intersection    of    cylinder 
and  plane,  114,  118 

normal  distribution,  224 

probability,  220,  223,  224 

sine,  93 

tangent,  94 

line  of  symmetry  of,  52 

point  of  symmetry  of,  52 

Decreasing  curve,  52 
Degree,  81 

Dependent  events,  210 
Depression,  angle  of,  86 
Deviation,  standard,  226,  227 
Difference,  tabular,  68 
angles  of  line,  243 


Difference,  cosines  of  line,  243 

negative,  24 

positive,  24 
Directions,  general,  for  preparing 

work,  5 

Directrix  of  parabola,  52 
Distance  between  two  points,  240 

focal,  136 
Division,  15 

with  slide  rule,  152 

Eccentricity  of  the  ellipse,  130 

of  the  circle,  136 
Elevation,  angle  of,  86 
Ellipse,  128 

axis  of,  133 

area  of,  136 

eccentricity  of,  134,  136 

equation  of,  129 

focal  distance  of,  136 

foci  of,  134 

general  shape  of,  130 

locating  points  upon,  133 

sum  of  focal  radii  of,  134 

tangent  to,  139 
Empirical,  equation,  47,  54,  57 

equations,  264,  266 
Equation,  of  circle,  37 

of  curve,  37 

of  ellipse,  129 

empirical,  47 

graph  of,  37 

linear,  44 

of  locus,  241,  243 

of  plane,  245 

solution  of,  by  factoring,  130 

of  straight  line,  40 

of  straight  line  through  two 
points,  45 

of  the  first  degree  between 

x  and  y,  43 
Equations,  conditional,  45 


INDEX 


305 


Equafions,  empirical,  264,  266 

of  equilibrium,  176 

plotted     upon     logarithmic 
paper,  281 

quadratic,  21 
Equilibrium,  equations  of,  176 

of  a  system  of  parallel  forces, 

171 

Error,  probable,  228 
Errors,  231 

of  area  of  triangle,  236,  237 

of  cube,  235 

of  cube  root,  235 

of  observation,  220 

of  product,  232 

of  quotient,  233 

of  sine,  235 

of  square,  233 

of  square  root,  234 

small,  231 
Events,  dependent,  210 

independent,  210 

mutually  exclusive,  209 
Expansion,   binomial,    189,    190, 

195 

Expectation,  212 
Exponents,  20 

Factoring,  16 

Falling    body,    distance    passed 

over  by,  51 
velocity  of,  51 
acceleration  of,  51 
Functions,  graph  of  trigonomet- 
ric, 92 
line    representation    of    the 

trigonometric,  94 
Focal  distance,  136    K 
radii,  sum  of,  134 
Foci  of  the  ellipse,  134 
Focus  of  the  ellipse,  134 
of  the  parabola,  52 
20 


Force,  concurrent,  159 

coplanar,  159 

moment  of,  165 
Forces,  in  equilibrium,  170 

polygon  of,  161 

resultant  of,  161 
Formulas,  addition,  108 

of  algebra,  288 

approximation,  196 

of  analytic  geometry,  292 

of  theory  of  probability,  295 

of  trigonometry,  289 

of  mensuration,  293 
Fractions,  18 
Frequency  distribution  polygon, 

222 

Function,  84 
Functions,  circular,  80,  84 

even,  90 

odd,  90 

trigonometric,  84 

of  0°,  90°,  180°,  and  270°,  87 

Grade,  percent,  42 

Gradient,  hydraulic,  42,  45 

Graph,  of  an  equation,  37 

of  an  equation   of  the  first 

degree  in  x  and  y,  43 
of  y  =  px2,  49 

Graphic    tables    of  the  trigono- 
metric functions,  96 

Graphs     of     the     trigonometric 
functions,  92 

Greek  alphabet,  286 

Head  of  water  in  well,  46 
Hyperbola,  equilateral,  23,  55 

Imaginary,  132 
Increasing  curve,  52 
Independent  events,  210 
Infinity,  88 


306 


INDEX 


Instruments,  1 
Intercept,  x-,  44 

y-,  42 

Law,  compound  interest,  66,  197 

of  cosines,  105 

empirical,  47 

of  freely  falling  body,  51 

of  sines,  104 

of  tangents,  106 
Least  squares,  227 

coefficients    determined    by, 

268 
Line  to  draw,  6 

direction  angles  of,  244 
cosines  of,  243 

equation  of,  40,  45 

projection  of,  244 

slope  of,  42 

straight,  23 

of  symmetry,  52 
Lines,  parallel,  8 
Locus,  equation  of,  241,  243 
Logarithm,  61 

base  of,  61 

characteristic  of,  64 

common,  66 

hyperbolic,  66 

Naperian,  66 

natural,  66 

mantissa  of,  64 

of  power,  64 

of  product,  62 

of  quotient,  63 

of  root,  64 
Logarithmic  curve,  73 

paper,  270,  277 

scales,  74,  147,  148 

tables,  67 

Materials,  1 
Magnitude,  angular,  80 


Mantissa  of  logarithm,  64 
Maxima  amd  minima,  252,  256, 

259 
Maximum    or      minimum       of 

ax2  +  bx  +  c,  255 
Mean,  arithmetic,  220,  228 
Measurements,  biological,  229 
Mensuration,  formulas  of,  293 
Meter,  current,  4 
Minima  (See  maxima) 
Minimum  (See  maximum) 
Minute,  81 
Moment,  168 

of  couple,  170 

of  force,  165 
Moments,  origin  of,  168 

sum  of,  174 

Multiple  logarithmic  paper,  277 
Multiplication,  13 

tables,  229 

with  slide  rule,  151 

Normal  cross-section  of  cylinder, 

115 
equation  of  plane,  245 

Octants,  239 
Ordinate,  25,  33 
Origin,  25,  239 

of  coordinates,  25 

of  moments,  168 
Orthographic  projection,  136 
of  circle,  138 

Paper,  coordinate,  25 

logarithmic,  270 

polar  coordinate,  4 

semi-logarithmic,  283 

squared,  25 
Parabola,  23,  52 

axis  of,  52 

directrix  of,  52 


INDEX 


307 


Parabola,  equation  of,  52 

focus  of,  52 

line  of  symmetry  of,  52 

property  of,  52 

vertex  of,  52 

Paraboloid  of  revolution,  53 
Parallel  forces,  165,  166,  170 
Pattern  for  circular  roof,  117 

for  right  circular  cylindrical 
surface  cut  by  a  plane, 
116 

for  saddle  of  ventilator,  78, 

115 

Pencil  points,  to  sharpen,  1 
Pencils,  4H,  1 

3H,  2 

chisel  pointed,  1,  2 

drawing,  1 

hexagonal,  2 
Permutations,  189 

formula  for  number  of,  192, 

193 
Plane,  slope  of,  247 

normal  equation  of,  245 
Planes,  coordinate,  239 
Plolling  statistical  data,  28 
Point,  33,  239 

abscissa  of,  33 

coordinates  of,  33 

ordinate  of,  33 

of  symmetry,  52 
Points,  distance  between,  240 

locating,  upon  ellipse,  133 
Polar  coordinate  paper,  4,  96 
Polygon,  force,  161 

frequency  distribution,  222, 

225 

Precipitation  charts,  30 
Precision,  measure  of,  226 
Pressure,   relation  between  vol- 
ume of  air  and,  58 


measured  in  inches  of  mer- 
cury and  feet  of  water, 
36 

Principle  of  slide  rule,  146 
Probability,  207 

curve,  223,  224 

theory  of,  formulas  for,  295 
Product,  error  of,  232 

logarithm  of,  62 
Progression,  arithmetical,  200 

geometrical,  200,  203 

infinite  geometrical,  205 
Projection,  of  line,  244 

orthographic,  136 
Proportional  parts,  tables  of,  70 
Protractor,  3 
Pulley,  rope  and,  185 

Quadrants,  34 
Quadratic,  general,  53 
Quotient,  error  of,  233 
logarithm  of,  63 

Radian,  91 
Radicals,  21 
Residual,  227 
Resultant,  force,  159,  161 

velocity,  158 
Revolution,  solid  of,  76 

paraboloid  of,  53 
Rock,  voids  of,  47 
Roof,  pattern  for  conical,  117 

volume  of  conical,  123 
Root,  logarithm  of,  64 
Rope  and  pulley,  185 
Rotation,  positive,  80 

negative,  80 

Saddle  for  ventilator,  paitern  for, 

78 

Sand-paper,  2 
Sand  voids  of,  47 
Scale,  3,  23 


308 


INDEX 


Scale  chain,  23 

double,  36,  74 

drawing  to,  23 

full  divided,  23 

logarithmic,  74 

non-uniform,  23,  74,  124 

open  divided,  23 

triangular,  4 
Scales,  double,  148 

logarithmic,  147,  148 
Second,  81 
Section  of  cone,  139 
Semi-logarithmic  paper,  283 
Sine  curve,  93 
Sines,  law  of,  104 
Slide  rule,  143,  150 

cube  roots  found  with,  153 

division  with,  152 

multiplication  with,  151 

principle  of,  146 

square  roots  found  with,  153 

squares  found  with,  153 
Slope,  equation  of  straight  line, 
42 

of  line,  42 

of  plane,  247 
Solid  of  revolution,  76 
Solution  of  triangle,  98 
Square,  error  of,  233 

root,  error  of,  234 

found  with  slide  rule,  153 

T-,  2 
Squares,    found  with  slide  rule, 

153 

Statics,  155 
Subtraction,  12 

vector,  156 
Sucessive  trials,  212 
Sum,  of  moments,  175 

vector,  156 
Symbols,  286 

of  aggregation,  16 


Symmetry,  line  of,  52 
point  of,  52 

Table  of  board  measure,  295 
converting  inches  into  feet 

295 

conversion,  298 
of    decimal    equivalents   of 
common  fractions,  295 
of  head  of  water  and  pres- 
sure, 297 
of      natural    trigonometric 

functions,  300 

of   water   contained   in    cir- 
cular pipe,  297 
Tables,  anti-logarithmic,  73 
logarithmic,  5,  68 
multiplication,  229 
of  proportional  parts,  70 
of   trigonometric   functions, 

96,  300 

Tangent  to  curve,  94 
to  ellipse,  139 
to  y  =  pxz,  50 
Tangents,  law  of,  106 
Temperature,   relative   between, 
and    density  of  water, 
54 

Tension,  179,  186 
Terms  of  (p  +  q)n,  215 
Translating  curve  on  logarithmic 

paper,  280 
Triangle,  80 

area  of,  71,  110,  111 
error  of  area  of,  236,  237 
drawing,  2 
of  reference,  84 
solution  of,  98 
60°,  2 
45°,  2 

Trigonometry,  formulas  of,  289 
Trials,  successive,  210 


INDEX 


309 


Unit,  circular,  91 

of  angular  measure,  81,  91 

Variation,  direct,  41 

inverse,  56 

Velocity  of  water  through  sand, 
46 

of  falling  body,  51 

curve,  54 

in  lower  Mississippi,  55 

resultant,  158 
Vector,  155 

addition,  157 

negative,  157 

sum,  157 

subtraction,  157 

Ventilator,  pattern  for  saddle  for, 
78 


Versed  sine,  95 
Vertex  of  angle,  80 

of  parabola,  52 
Voids  of  rock,  47 

of  sand,  47 
Volume  of  air  under  pressure,  56 

of  conical  roof,  123 

of  right  circular  cylindrical 
shell,  120 

of  solid  of  revolution,  77 

Water,  relation  between  tempera- 
ture and  density  of,  54 
velocity  of,  through  sand,  46 
Well,  yield  of,  46 
Work,     arrangement     of     loga- 
rithmic, 71,  74 
blocking  out  of,  71 


